Solution:
Let $k$ be one of the elements in Alices set $A$ of positive integers. The number of sets that Bob lists with the property that their maximum element is k is $2^{k-1}$, since every positive integer less than k can be in the set or out. Thus, for the number of sets bob have listed to be 2024, we want to find a sum of unique powers of two that can achieve this. 2024 is equal to $2^{10}+2^9+2^8+2^7+2^6+2^5+2^3$. We must increase each power by 1 to find the elements in set $A$, which are $(11,10,9,8,7,6,4)$. Add these up to get $\boxed{055}$. -westwoodmonster
Let $A = \left\{ a_1, a_2, \cdots, a_n \right\}$ with $a_1 < a_2 < \cdots < a_n$.
If the maximum element of $B$ is $a_i$ for some $i \in \left\{ 1, 2, \cdots , n \right\}$, then each element in $\left\{ 1, 2, \cdots, a_i- 1 \right\}$ can be either in $B$ or not in $B$.
Therefore, the number of such sets $B$ is $2^{a_i - 1}$.
Therefore, the total number of sets $B$ is
\begin{align*}
\sum_{i=1}^n 2^{a_i - 1} & = 2024 .
\end{align*}
Thus
\begin{align*}
\sum_{i=1}^n 2^{a_i} & = 4048 .
\end{align*}
Now, the problem becomes writing 4048 in base 2, say, $4048 = \left( \cdots b_2b_1b_0 \right)_2$.
We have $A = \left\{ j \geq 1: b_j = 1 \right\}$.
We have $4048 = \left( 111,111,010,000 \right)_2$.
Therefore, $A = \left\{ 4, 6, 7, 8, 9, 10, 11 \right\}$.
Therefore, the sum of all elements in $A$ is $\boxed{\textbf{(55) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Answer: 055

<｜begin▁of▁sentence｜><｜User｜>Alice chooses a set $A$ of positive integers. Then Bob lists all finite nonempty sets $B$ of positive integers with the property that the maximum element of $B$ belongs to $A$. Bob's list has 2024 sets. Find the sum of the elements of A.<｜Assistant｜><think>

Okay, so Alice picks a set A of positive integers, and then Bob lists all finite nonempty sets B where the maximum element of B is in A. The total number of such sets B is 2024. We need to find the sum of the elements in A. Hmm, let me try to break this down step by step.

First, let me understand the problem again. Alice chooses some set A, which is a set of positive integers. Then Bob has to list all possible finite non-empty sets B such that the maximum element in each B is an element of A. The total number of these sets B is given as 2024. Our job is to figure out what A must be so that this condition holds, and then compute the sum of the elements in A.

Alright, so maybe I should start by thinking about how the number of sets B relates to the set A. Let me consider smaller examples first to see if I can find a pattern or formula.

Suppose A is a singleton set, say A = {a}. Then, what would be the number of sets B that Bob lists? Well, any non-empty set B where the maximum element is a. So, the elements of B can be any non-empty subset of {1, 2, ..., a} that includes a. The number of such subsets is 2^{a-1}, since for each element less than a, we can choose to include it or not, and we must include a. So, the total is 2^{a-1}.

Therefore, if A has only one element a, then the number of sets B is 2^{a-1}. So, if 2^{a-1} = 2024, then a would be log2(2024) + 1. But 2^11 = 2048, so 2^11 is 2048, so 2^10 is 1024. 2024 is between 2^10 and 2^11. Since 2024 isn't a power of two, this suggests that A must have more than one element.

Hmm, so maybe A has multiple elements. Let me think about how the number of sets B increases when A has multiple elements. Let's suppose A has two elements, say A = {a, b} where a < b. Then, the sets B that Bob lists are all non-empty finite sets where the maximum element is either a or b.

To compute the total number of such sets, we can consider the sets where the maximum is a, plus those where the maximum is b. The number of sets with maximum a is 2^{a-1}, as before. The number of sets with maximum b is 2^{b-1}. However, we have to be careful here: if a < b, then the sets with maximum b can include any elements up to b, but they must include b and cannot have any elements larger than b. However, since a is in A as well, but a < b, the sets with maximum a are subsets where all elements are ≤ a, and the maximum is exactly a. Similarly for b.

Therefore, the total number of sets B would be 2^{a-1} + 2^{b-1}. Wait, but is that correct? Let me check with an example. Let's take A = {1, 2}. Then the sets B would be:

- All non-empty subsets of {1,2} where the maximum is 1 or 2.

The maximum can be 1 or 2. The sets with maximum 1 are just {1}. The sets with maximum 2 are {2}, {1,2}. So total sets: 1 + 2 = 3.

According to the formula, 2^{1-1} + 2^{2-1} = 1 + 2 = 3. That works. Another example: A = {2,3}. Then the sets with maximum 2 are {2}, and those with maximum 3 are {3}, {1,3}, {2,3}, {1,2,3}. Wait, hold on. Wait, if the maximum is 3, then the elements can be any subset of {1,2,3} that includes 3 and has maximum 3. So the subsets are all subsets of {1,2,3} that include 3 and are non-empty. The number is 2^{2} = 4. Similarly, the sets with maximum 2 are subsets of {1,2} that include 2, so 2^{1} = 2. Wait, but in my previous example with A={2,3}, the total would be 2 + 4 = 6. Let me list them:

Maximum 2: {2}, {1,2}

Maximum 3: {3}, {1,3}, {2,3}, {1,2,3}

Total 6 sets. So the formula holds. So yes, for A = {a, b}, with a < b, the total number is 2^{a-1} + 2^{b-1}.

Wait, but in the first example with A={1,2}, the total was 3, which is 1 + 2. So that works.

Therefore, in general, if A is a set with elements a_1 < a_2 < ... < a_n, then the total number of sets B is the sum over each a_i of 2^{a_i - 1}, but subtracting something? Wait, no, actually, wait. Wait, actually, if the elements of A are not consecutive, does that affect the count?

Wait, let me think again. Suppose A has elements that are not consecutive. Let's take A = {1,3}. Then the sets B with maximum in A would be:

Maximum 1: {1}

Maximum 3: All subsets of {1,2,3} that include 3 and have maximum 3. So the subsets are {3}, {1,3}, {2,3}, {1,2,3}. So that's 4 sets. So total is 1 + 4 = 5.

According to the formula, 2^{1-1} + 2^{3-1} = 1 + 4 = 5. So that works. So even if there are gaps in A, the formula still holds. Because when considering the maximum element a_i, the elements below can be any subset of the numbers less than a_i, but since the maximum is a_i, they can include any numbers less than a_i, regardless of whether those numbers are in A or not.

Therefore, the formula is indeed, for each element a in A, the number of subsets where the maximum is a is 2^{a-1}, so the total number of sets B is the sum over all a in A of 2^{a-1}.

Therefore, the problem reduces to finding a set A of positive integers such that the sum of 2^{a-1} over all a in A equals 2024, and then finding the sum of the elements of A.

So, we need to find a set A where sum_{a in A} 2^{a-1} = 2024, and then compute sum_{a in A} a.

Therefore, the key is to express 2024 as a sum of distinct powers of 2, but shifted by exponents (a-1). Wait, but 2^{a-1} for different a's are distinct powers of 2 only if the a's are distinct. Since A is a set, the a's are distinct, so each term 2^{a-1} is a distinct power of 2. Wait, actually, 2^{a-1} for different a's are different powers of two. For example, if a=1, it's 1 (2^0), a=2 is 2 (2^1), a=3 is 4 (2^2), etc. So each term is a power of two, and since the a's are distinct, each exponent is unique. Therefore, the sum is just a sum of distinct powers of two. But any number can be expressed as a sum of distinct powers of two, that's binary representation.

Wait, but hold on. The problem is that the exponents here are (a-1), so each term is 2^{a-1}, so each term corresponds to a power of two where the exponent is one less than the element in A. Therefore, the sum over 2^{a-1} is equal to 2024. So, if we write 2024 in binary, each bit set in the binary representation corresponds to a term 2^{k}, which would mean that (a-1) = k, so a = k+1. Therefore, the elements of A are exactly the numbers (k+1) where the k-th bit (starting from 0) is set in the binary representation of 2024.

Therefore, to find A, we can write 2024 in binary, and then each bit that is set corresponds to an element in A. For example, if 2024 in binary is, say, 11111101000, then the positions of the 1s would correspond to exponents k, so the elements of A would be k+1 for each of those positions.

Therefore, first, let me compute the binary representation of 2024. Let me do that.

2024 divided by 2: 2024 /2 = 1012, remainder 0

1012 /2 = 506, rem 0

506 /2 = 253, rem 0

253 /2 = 126 rem 1

126 /2 = 63 rem 0

63 /2 = 31 rem 1

31 /2 =15 rem 1

15 /2 =7 rem 1

7/2=3 rem 1

3/2=1 rem 1

1/2=0 rem 1

So writing the remainders from last to first:

1 1 1 1 1 1 0 1 0 0 0

Wait let me count the divisions:

Starting from 2024:

1. 2024 /2 = 1012, rem 0 (LSB)

2. 1012 /2 = 506, rem 0

3. 506 /2 =253, rem 0

4. 253 /2=126 rem 1

5. 126 /2=63 rem 0

6. 63 /2=31 rem 1

7. 31 /2=15 rem 1

8. 15 /2=7 rem 1

9. 7 /2=3 rem 1

10. 3 /2=1 rem 1

11. 1 /2=0 rem 1 (MSB)

So the remainders from first to last (LSB to MSB) are:

0,0,0,1,0,1,1,1,1,1,1

But when written as binary, we take the remainders from last division to first, so the MSB first:

The first remainder is the LSB, so the binary is written as the remainders in reverse order:

The last division gave remainder 1 (step 11), then step 10 gave 1, step 9 gave 1, step 8 gave 1, step7 gave 1, step6 gave 1, step5 gave 0, step4 gave 1, step3 gave 0, step2 gave 0, step1 gave 0.

Wait, perhaps I need to list them in reverse order. Let me list the remainders in the order they were obtained, from first to last (which is LSB to MSB):

First division (step1): remainder 0 (LSB)

step2: 0

step3:0

step4:1

step5:0

step6:1

step7:1

step8:1

step9:1

step10:1

step11:1 (MSB)

Therefore, the binary digits from LSB to MSB are:

0 (2^0), 0 (2^1), 0 (2^2), 1 (2^3), 0 (2^4), 1 (2^5), 1 (2^6), 1 (2^7), 1 (2^8), 1 (2^9), 1 (2^10)

Therefore, the binary representation is the MSB first, so starting from the last remainder:

The MSB is the 11th remainder (step11) which is 1 (2^10), then step10:1 (2^9), step9:1 (2^8), step8:1 (2^7), step7:1 (2^6), step6:1 (2^5), step5:0 (2^4), step4:1 (2^3), step3:0 (2^2), step2:0 (2^1), step1:0 (2^0).

Therefore, writing the binary number from MSB to LSB:

1 1 1 1 1 1 0 1 0 0 0

So that's 11 bits. Let me check:

2^10 is 1024, so 2^10 + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 0 + 2^3 + 0 + 0 + 0.

Wait, let's compute that:

2^10 = 1024

2^9 = 512, total so far 1536

2^8 = 256, total 1792

2^7 = 128, total 1920

2^6 = 64, total 1984

2^5 = 32, total 2016

Then 2^3 = 8, so total 2016 +8 = 2024. Yes, that works. So the binary is 11111101000.

So the binary digits are positions 10 down to 0:

Position 10:1, 9:1,8:1,7:1,6:1,5:1,4:0,3:1,2:0,1:0,0:0.

Therefore, the bits set are at positions 10,9,8,7,6,5,3.

Therefore, each set bit corresponds to an exponent k in 2^k, so the elements of A are (k+1) for each set bit.

Therefore, the elements of A are:

For position 10: k=10 → a=11,

position9: a=10,

position8: a=9,

position7: a=8,

position6: a=7,

position5: a=6,

position3: a=4.

Wait, position 3 is 2^3, so k=3, so a=4.

Wait, so the elements are 11,10,9,8,7,6,4.

Wait, let me list them all:

Bits set at positions 10,9,8,7,6,5,3.

So:

Position 10: a =10 +1=11,

Position9: 9+1=10,

Position8:8+1=9,

Position7:7+1=8,

Position6:6+1=7,

Position5:5+1=6,

Position3:3+1=4.

So the set A is {4,6,7,8,9,10,11}.

Therefore, the elements are 4,6,7,8,9,10,11.

Therefore, the sum of elements in A is 4 +6 +7 +8 +9 +10 +11. Let me compute that:

4+6=10,

10+7=17,

17+8=25,

25+9=34,

34+10=44,

44+11=55.

So the total sum is 55.

Wait, but let me confirm that this is correct. Let me check again.

First, the binary representation of 2024 is indeed 11111101000. Let me confirm that:

Let me compute 2^10 + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^3.

2^10 = 1024,

2^9 = 512,

2^8 = 256,

2^7 = 128,

2^6 = 64,

2^5 = 32,

2^3 = 8.

Adding them up:

1024 + 512 = 1536,

1536 +256=1792,

1792+128=1920,

1920+64=1984,

1984+32=2016,

2016 +8=2024.

Yes, that's correct.

Therefore, the exponents k where the bits are set are 10,9,8,7,6,5,3. Therefore, the elements a are k+1, so 11,10,9,8,7,6,4.

Therefore, the set A is {4,6,7,8,9,10,11}, and the sum is indeed 4+6+7+8+9+10+11. Let me add them again step by step:

4 + 6 = 10,

10 +7 =17,

17+8=25,

25+9=34,

34+10=44,

44+11=55.

Yes, 55.

Therefore, the answer should be 55.

But wait, let me make sure that there isn't another possible set A that could also sum to 2024. Since the problem states that Bob's list has 2024 sets, and the number of sets is the sum of 2^{a-1} over A. Since each term 2^{a-1} is a power of two, the representation of 2024 as a sum of distinct powers of two is unique (since binary representation is unique). Therefore, the set A must be exactly the set {4,6,7,8,9,10,11}, so the sum is 55.

Therefore, I think the answer is 55.

Wait, but let me check with another approach to be safe.

Suppose A is the set {4,6,7,8,9,10,11}. Let's compute the total number of sets B.

For each a in A, compute 2^{a-1} and sum them up:

For 4: 2^3 =8,

6:2^5=32,

7:2^6=64,

8:2^7=128,

9:2^8=256,

10:2^9=512,

11:2^{10}=1024.

Adding these up:

8 +32 =40,

40+64=104,

104+128=232,

232+256=488,

488+512=1000,

1000+1024=2024.

Yes, that's exactly the number given. So that checks out.

Therefore, the sum of elements in A is indeed 55.

Therefore, the answer is 55.

**Final Answer**
\boxed{55}
</think>

Alice chooses a set \( A \) of positive integers. Bob lists all finite nonempty sets \( B \) of positive integers such that the maximum element of \( B \) belongs to \( A \). The number of such sets \( B \) is 2024. We need to find the sum of the elements of \( A \).

To solve this, we note that the number of sets \( B \) with maximum element \( a \) is \( 2^{a-1} \). Therefore, the total number of sets \( B \) is the sum of \( 2^{a-1} \) for all \( a \) in \( A \). This sum must equal 2024.

We convert 2024 to its binary representation to identify the elements of \( A \). The binary representation of 2024 is \( 11111101000_2 \), which corresponds to the powers of 2:

\[
2024 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^3
\]

Each term \( 2^{k} \) corresponds to an element \( a = k + 1 \) in \( A \). Thus, the elements of \( A \) are:

\[
11, 10, 9, 8, 7, 6, 4
\]

Summing these elements:

\[
4 + 6 + 7 + 8 + 9 + 10 + 11 = 55
\]

Thus, the sum of the elements of \( A \) is \(\boxed{55}\).


Solution:
We note that by changing a digit to $1$ for the number $\overline{abcd}$, we are subtracting the number by either $1000(a-1)$, $100(b-1)$, $10(c-1)$, or $d-1$. Thus, $1000a + 100b + 10c + d \equiv 1000(a-1) \equiv 100(b-1) \equiv 10(c-1) \equiv d-1 \pmod{7}$. We can casework on $a$ backwards, finding the maximum value.
(Note that computing $1000 \equiv 6 \pmod{7}, 100 \equiv 2 \pmod{7}, 10 \equiv 3 \pmod{7}$ greatly simplifies computation).
Applying casework on $a$, we can eventually obtain a working value of $\overline{abcd} = 5694 \implies \boxed{699}$. ~akliu
Let our four digit number be $abcd$. Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Reducing, we get
$6+2b+3c+d \equiv 0 \pmod{7}$ $(1)$
$6a+2+3c+d \equiv 0 \pmod{7}$ $(2)$
$6a+2b+3+d \equiv 0 \pmod{7}$ $(3)$
$6a+2b+3c+1 \equiv 0 \pmod{7}$ $(4)$
Subtracting $(2)-(1), (3)-(2), (4)-(3), (4)-(1)$, we get:
$3a-b \equiv 2 \pmod{7}$
$2b-3c \equiv 6 \pmod{7}$
$3c-d \equiv 2 \pmod{7}$
$6a-d \equiv 5 \pmod{7}$
For the largest 4 digit number, we test values for a starting with 9. When a is 9, b is 4, c is 3, and d is 7.  However, when switching the digits with 1, we quickly notice this doesnt work. Once we get to a=5, we get b=6,c=9,and d=4. Adding 694 with 5, we get $\boxed{699}$ -westwoodmonster
Let our four digit number be $abcd$. Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Add the equations together, we get:
$3000a+300b+30c+3d+1111 \equiv 0 \pmod{7}$
And since the remainder of 1111 divided by 7 is 5, we get:
$3abcd \equiv 2 \pmod{7}$
Which gives us:
$abcd \equiv 3 \pmod{7}$
And since we know that changing each digit into 1 will make abcd divisible by 7, we get that $d-1$, $10c-10$, $100b-100$, and $1000a-1000$ all have a remainder of 3 when divided by 7. Thus, we get $a=5$, $b=6$, $c=9$, and $d=4$. Thus, we get 5694 as abcd, and the answer is $694+5=\boxed{699}$.
~Callisto531
Let our four digit number be $abcd$. Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Then, we let x, y, z, t be the smallest whole number satisfying the following equations:
$1000a \equiv x \pmod{7}$
$100b \equiv y \pmod{7}$
$10a \equiv z \pmod{7}$
$d \equiv t \pmod{7}$
Since 1000, 100, 10, and 1 have a remainder of 6, 2, 3, and 1 when divided by 7, we can get the equations of:
(1): $6+y+z+t \equiv 0 \pmod{7}$
(2): $x+2+z+t \equiv 0 \pmod{7}$
(3): $x+y+3+t \equiv 0 \pmod{7}$
(4): $x+y+z+1 \equiv 0 \pmod{7}$
Add (1), (2), (3) together, we get:
$2x+2y+2z+3t+11 \equiv 0 \pmod{7}$
We can transform this equation to:
$2(x+y+z+1)+3t+9 \equiv 0 \pmod{7}$
Since, according to (4), $x+y+z+1$ has a remainder of 0 when divided by 7, we get:
$3t+9 \equiv 0 \pmod{7}$
And because t is 0 to 6 due to it being a remainder when divided by 7, we use casework and determine that t is 4.
Using the same methods of simplification, we get that x=2, y=5, and z=6, which means that 1000a, 100b, 10c, and d has a remainder of 2, 5, 6, and 4, respectively. Since a, b, c, and d is the largest possible number between 0 to 9, we use casework to determine the answer is a=5, b=6, c=9, and d=4, which gives us an answer of $5+694=\boxed{699}$
~Callisto531 and his dad

Answer: 699

<｜begin▁of▁sentence｜><｜User｜>Let $N$ be the greatest four-digit positive integer with the property that whenever one of its digits is changed to $1$, the resulting number is divisible by $7$. Let $Q$ and $R$ be the quotient and remainder, respectively, when $N$ is divided by $1000$. Find $Q+R$.<｜Assistant｜><think>

Okay, so I need to find the greatest four-digit number N such that if any one of its digits is changed to 1, the resulting number is divisible by 7. Then, when N is divided by 1000, I have to find the sum of the quotient Q and the remainder R. Hmm, let's break this down step by step.

First, let me understand the problem clearly. N is a four-digit number, so it ranges from 1000 to 9999. We need the largest such number where changing any single digit to 1 results in a number divisible by 7. That means, for each digit in N (thousands, hundreds, tens, units), if I replace that digit with 1 (keeping the others the same), the new number must be divisible by 7.

So, for example, if N is ABCD (where A, B, C, D are digits), then changing A to 1 gives 1BCD, changing B to 1 gives A1CD, changing C to 1 gives AB1D, and changing D to 1 gives ABC1. Each of these four numbers must be divisible by 7.

Our goal is to find the maximum such N, then compute Q + R where N = 1000*Q + R. Since dividing by 1000, Q is the first digit (thousands place) and R is the last three digits. So Q + R would be A + BCD (as a number). Wait, actually, if N is ABCD, then Q is A and R is BCD, so Q + R is A + BCD. But since BCD is a three-digit number, adding A (a single digit) to it would just be the number formed by A followed by BCD? Wait no, no. Wait, Q is the quotient when N is divided by 1000, so Q is floor(N / 1000), which is the thousands digit. R is the remainder, which is N mod 1000, so the last three digits. So Q is A, R is BCD, so Q + R is A + BCD (as numbers). For example, if N is 1234, then Q is 1, R is 234, so Q + R is 1 + 234 = 235. Got it.

So first, need to find N, the largest four-digit number with that divisibility condition. Let me think how to approach this.

Since we need the largest such N, we can start from 9999 and check downwards, but that might be too time-consuming. Instead, perhaps we can set up equations based on the divisibility conditions.

Let me denote the four-digit number as N = 1000a + 100b + 10c + d, where a, b, c, d are digits from 0-9, except a which is from 1-9.

The conditions are:

1. Changing the thousands digit to 1: 1000*1 + 100b + 10c + d must be divisible by 7.
2. Changing the hundreds digit to 1: 1000a + 100*1 + 10c + d must be divisible by 7.
3. Changing the tens digit to 1: 1000a + 100b + 10*1 + d must be divisible by 7.
4. Changing the units digit to 1: 1000a + 100b + 10c + 1 must be divisible by 7.

So each of these four numbers must be divisible by 7. Let me write these as equations modulo 7.

Let me denote N as 1000a + 100b + 10c + d. Let me compute each modified number modulo 7.

First, let me note that 1000 ≡ -1 mod 7 because 1001 is divisible by 7 (since 7*143=1001). Similarly, 100 ≡ 2 mod 7 (since 100 - 14*7=100-98=2), 10 ≡ 3 mod7, and 1≡1 mod7. Wait, let me confirm these:

Compute 10 mod7 =3, 10^2 mod7= 3*3=9 mod7=2, 10^3 mod7= 2*3=6 mod7, 10^4 mod7=6*3=18 mod7=4, but maybe I need to compute each coefficient:

Wait, let me compute 1000 mod7:

1000 divided by 7: 7*142=994, so 1000-994=6. So 1000 ≡6 mod7. Wait, but 10^3 mod7:

10 mod7=3, 10^2 mod7= 3*3=9 mod7=2, 10^3 mod7= 2*3=6 mod7, yes, so 10^3 ≡6 mod7. Similarly, 100 is 10^2 ≡2 mod7, 10^1≡3 mod7, 10^0≡1 mod7.

Therefore, the number N = a*1000 + b*100 + c*10 + d ≡ a*6 + b*2 + c*3 + d*1 mod7.

Similarly, when we change each digit to 1, let's compute each modified number modulo7.

First, changing the thousands digit to 1:

Number becomes 1*1000 + b*100 + c*10 + d. Its mod7 is 1*6 + b*2 + c*3 + d*1 ≡6 + 2b +3c +d mod7. This must be ≡0 mod7.

Second, changing hundreds digit to1:

a*1000 +1*100 +c*10 +d. Mod7: 6a +1*2 +3c +d ≡6a +2 +3c +d mod7 ≡0 mod7.

Third, changing tens digit to1:

a*1000 +b*100 +1*10 +d. Mod7:6a +2b +3*1 +d ≡6a +2b +3 +d mod7 ≡0 mod7.

Fourth, changing units digit to1:

a*1000 +b*100 +c*10 +1. Mod7:6a +2b +3c +1 mod7 ≡0 mod7.

Therefore, we have four congruences:

1. 6 + 2b + 3c + d ≡0 mod7. (Changing thousands digit)
2. 6a + 2 + 3c + d ≡0 mod7. (Changing hundreds digit)
3. 6a + 2b + 3 + d ≡0 mod7. (Changing tens digit)
4. 6a + 2b + 3c +1 ≡0 mod7. (Changing units digit)

Additionally, N itself is 1000a + 100b +10c +d, and we need the maximum such N, so we want to maximize a, then b, then c, then d.

So perhaps we can set up these equations and solve for a,b,c,d.

Let me write the four equations again:

Equation1: 2b + 3c + d ≡ -6 mod7 → 2b +3c +d ≡1 mod7 (since -6 mod7 is 1)

Equation2:6a +2 +3c +d ≡0 mod7 →6a +3c +d ≡-2 mod7 →6a +3c +d ≡5 mod7 (since -2 mod7 is 5)

Equation3:6a +2b +d +3 ≡0 mod7 →6a +2b +d ≡-3 mod7 →6a +2b +d ≡4 mod7 (since -3 mod7 is 4)

Equation4:6a +2b +3c +1 ≡0 mod7 →6a +2b +3c ≡-1 mod7 →6a +2b +3c ≡6 mod7 (since -1 mod7 is 6)

So now, equations:

1. 2b +3c +d ≡1 mod7

2.6a +3c +d ≡5 mod7

3.6a +2b +d ≡4 mod7

4.6a +2b +3c ≡6 mod7

Hmm, perhaps we can subtract equations to eliminate variables.

Let me see equations 1 and 2. Let's subtract equation1 from equation2:

(6a +3c +d) - (2b +3c +d) ≡5 -1 mod7 →6a -2b ≡4 mod7 →6a -2b ≡4 mod7.

Similarly, equation3 minus equation1:

(6a +2b +d) - (2b +3c +d) =6a -3c ≡4 -1 mod7 →6a -3c ≡3 mod7.

Equation4 is 6a +2b +3c ≡6 mod7.

Hmm, perhaps we can express variables in terms of others.

Alternatively, let me see equation4 and equation3.

Wait, equation4: 6a +2b +3c ≡6 mod7.

Equation3:6a +2b +d ≡4 mod7.

Subtract equation3 from equation4: (6a +2b +3c) - (6a +2b +d) ≡6 -4 mod7 →3c -d ≡2 mod7 →d ≡3c -2 mod7.

So from this, d ≡3c -2 mod7. Let me note that as equation5.

Similarly, equation1:2b +3c +d ≡1 mod7. Substitute d from equation5 into equation1:

2b +3c + (3c -2) ≡1 mod7 →2b +6c -2 ≡1 mod7 →2b +6c ≡3 mod7.

Equation6: 2b +6c ≡3 mod7.

Equation2:6a +3c +d ≡5 mod7. Again substitute d from equation5:

6a +3c + (3c -2) ≡5 mod7 →6a +6c -2 ≡5 mod7 →6a +6c ≡7 mod7 →6a +6c ≡0 mod7 (since 7 mod7 is 0). Wait, 6a +6c ≡5 +2 mod7? Wait, let me recalculate:

Wait equation2 substitution:

6a +3c + d =6a +3c + (3c -2 mod7). Wait, actually, d ≡3c -2 mod7, so substituting into equation2:

6a +3c + d ≡6a +3c + (3c -2) mod7 ≡6a +6c -2 mod7. This must be ≡5 mod7. Therefore:

6a +6c -2 ≡5 mod7 →6a +6c ≡5 +2 mod7 →6a +6c ≡7 mod7 →6a +6c ≡0 mod7 (since 7 mod7 is 0). So 6(a + c) ≡0 mod7. Since 6 and 7 are coprime, this implies (a + c) ≡0 mod7. So equation7: a + c ≡0 mod7.

Similarly, equation3 minus equation1 gave us equation3 - equation1: 6a -3c ≡3 mod7. Let me write that as equation8:6a -3c ≡3 mod7.

So equation7: a + c ≡0 mod7 → a ≡ -c mod7.

Equation8:6a -3c ≡3 mod7. Let me substitute a ≡-c mod7 into equation8:

6*(-c) -3c ≡3 mod7 →-6c -3c ≡3 mod7 →-9c ≡3 mod7 →-9c mod7: since -9 mod7 is -2 mod7 (because 9 mod7 is 2, so -9 mod7 is -2 mod7, which is 5 mod7). Wait, let me compute:

-9c mod7: 9 mod7 is 2, so -9c mod7 is (-2c) mod7. So:

-2c ≡3 mod7 →-2c ≡3 mod7 →Multiply both sides by -1: 2c ≡-3 mod7 →2c ≡4 mod7 (since -3 mod7 is 4). Therefore, 2c ≡4 mod7 →Divide both sides by 2 (since 2 and 7 coprime, inverse of 2 mod7 is 4 because 2*4=8≡1 mod7). So multiply both sides by 4:

c ≡4*4 mod7 →Wait, no, 2c ≡4 mod7 →c ≡2 mod7 (since 2*2=4). So c ≡2 mod7. So c can be 2 or 9 (since c is a digit from 0-9). Wait, 2 mod7 is 2, so possible c values are 2, 9 (since 2 +7=9, next would be 16 which is invalid). So c is either 2 or 9.

So c ∈ {2,9}. Let me note that.

From equation7: a ≡ -c mod7. Since a is a digit from 1-9 (as it's the thousands place), so:

If c=2, then a ≡-2 mod7 →a ≡5 mod7 (since -2 mod7 is 5). So possible a values are 5 or 12, but a is a single digit, so a=5 or 12 is invalid. So a=5.

If c=9, then a ≡-9 mod7. Since -9 mod7 is (-9 +14)=5 mod7. Wait, -9 mod7: 9 divided by7 is 1 with remainder 2, so -9 mod7 is -2 mod7 which is 5 mod7. Wait, same as c=2? Wait, c=9: -c mod7 is -9 mod7. Let me compute:

-9 mod7: 7*2=14, so -9 +14=5. So yes, -9 mod7 is 5. So a ≡5 mod7. So same as before. So a can be 5 or 12, but again a is a digit, so a=5 or a=5+7=12 which is invalid. So a=5. Wait, so regardless of c being 2 or 9, a must be 5? Hmm, that's interesting.

Wait, so if c is 2 or 9, then a must be 5. Wait, but let me check:

Wait, if c=2, then a ≡-2 mod7 →5 mod7, so a=5.

If c=9, then a ≡-9 mod7 ≡-2 mod7 ≡5 mod7, so a=5. So in both cases, a=5. So a is fixed at 5. Interesting.

So a=5. So now, c can be 2 or 9. Let's consider both cases.

First, let's consider c=2:

Case 1: c=2, a=5.

Now, let's see equation5: d ≡3c -2 mod7. Since c=2, d ≡6 -2=4 mod7. So d ≡4 mod7. So possible d digits are 4 or 11, but since d is a digit, d can be 4 or 11-7=4? Wait, 4 mod7 is 4, so d can be 4 or 11, but 11 is invalid. So d=4 or 11? Wait, no, mod7, so possible d values are 4, 11, 18,... but since d is a single digit (0-9), so d can be 4 or 11-7=4? Wait, no, 4 mod7 is 4, so the possible d digits are 4 and 4+7=11 which is invalid. So only d=4. Wait, but 3c -2 mod7: 3*2 -2 =6-2=4 mod7, so yes, d must be 4 mod7, so d=4 or 11, but 11 is invalid, so d=4.

Wait, but maybe I made a mistake here. Let me check equation5 again. equation5 was d ≡3c -2 mod7. So for c=2, that's 3*2 -2 =6-2=4 mod7, so d ≡4 mod7. So possible d digits are 4 or 11 (but 11 is invalid), so d must be 4. So d=4.

Now, let's look at equation6: 2b +6c ≡3 mod7. Since c=2, 6c=12, so 2b +12 ≡3 mod7 →2b +12 mod7: 12 mod7 is 5, so 2b +5 ≡3 mod7 →2b ≡-2 mod7 →2b ≡5 mod7 (since -2 mod7 is 5). Then, solving for b:

2b ≡5 mod7. Multiply both sides by inverse of 2 mod7, which is 4, since 2*4=8≡1 mod7. So b ≡5*4 mod7 →20 mod7=20-14=6 mod7. So b≡6 mod7. So possible b digits are 6 or 13, but since b is a digit (0-9), so b=6 or 13-7=6? Wait, 6 mod7 is 6, so possible b=6 or 13 (invalid). So b=6.

So in this case, with c=2, we have a=5, b=6, c=2, d=4. So the number is 5 6 2 4, i.e., 5624. Let me check if this satisfies all conditions.

Wait, but before that, let's check the other case where c=9.

Case 2: c=9, a=5.

Again, equation5: d ≡3c -2 mod7. 3*9=27, 27-2=25. 25 mod7: 25 divided by7 is 3*7=21, so 25-21=4 mod7. So d≡4 mod7. So same as before, d=4 or 11, so d=4.

Wait, so d=4 again. Hmm.

Now equation6: 2b +6c ≡3 mod7. Here c=9, so 6c=54. 54 mod7: 7*7=49, so 54-49=5 mod7. So equation6 becomes 2b +5 ≡3 mod7 →2b ≡-2 mod7 → same as before, 2b≡5 mod7 →b≡6 mod7. So again, b=6 or 13, so b=6.

Wait, so in this case, c=9, so the number would be a=5, b=6, c=9, d=4 →5694.

Wait, so in both cases c=2 and c=9, we get b=6, d=4, a=5, so numbers 5624 and 5694. But since we are to find the greatest such N, 5694 is larger than 5624, so perhaps c=9 is better. But we need to check if all the conditions are satisfied for both numbers. Let me check both.

First, let's check 5624:

Changing each digit to 1:

1. Change thousands digit to1: 1624. Check divisibility by7: 1624 ÷7. Let's compute 7*232=1624. 7*200=1400, 7*32=224, so 1400+224=1624. So yes, divisible by7.

2. Change hundreds digit to1: 5124. 5124 ÷7. Let's see 7*700=4900, 5124-4900=224. 224 ÷7=32. So total 732. 7*732=5124. So yes.

3. Change tens digit to1: 5614. 5614 ÷7. 7*800=5600, so 5614-5600=14, which is 2*7. So 802. 7*802=5614. Correct.

4. Change units digit to1: 5621. Hmm, 5621 ÷7. Let's compute 7*800=5600, 5621-5600=21. 21 ÷7=3. So total 803. 7*803=5621. Yes, divisible by7. So all four conditions are satisfied. So 5624 is valid.

Now check 5694:

Changing digits:

1. Change thousands digit to1: 1694. Check divisibility by7. Let's compute 1694 ÷7. 7*242=1694 (since 7*240=1680, plus 7*2=14, so 1680+14=1694). So yes.

2. Change hundreds digit to1: 5194. 5194 ÷7. Let's see 7*700=4900, 5194-4900=294. 294 ÷7=42. So total 742. 7*742=5194. Correct.

3. Change tens digit to1: 5614. Wait, same as before? Wait, original number is 5694. Changing tens digit (which is 9) to1 gives 5614. Wait, same as the third case for 5624. 5614 is divisible by7 as before. So that's okay.

4. Change units digit to1: 5691. Check 5691 ÷7. Let's compute 7*813=5691 (7*800=5600, 7*13=91, so 5600+91=5691). So yes, divisible by7.

Therefore, both 5624 and 5694 satisfy the conditions. Since we need the greatest such N, 5694 is larger, so that's better. But wait, maybe there are even larger numbers? Because we considered c=2 and c=9, but perhaps there are other possibilities?

Wait, hold on. Wait, in our analysis above, when c=9, we had a=5, but maybe there are other possibilities? Let me check again.

Wait, in equation7, a + c ≡0 mod7. So a = ( -c ) mod7. Since a is a digit from1-9, and c is a digit from0-9. So for c=9, a must be 5 mod7 (since -9 mod7 is 5). So a=5 or 12, but 12 is invalid, so a=5. So that's correct.

But perhaps there are other values of c? Wait, earlier when solving equation8, we found that c must be ≡2 mod7, so c can be 2 or 9. Since c is a digit, 0-9, so 2 and 9. So those are the only possibilities. So only c=2 and 9. So the only possible numbers are 5624 and 5694. Since 5694 is larger, that's the candidate. But wait, maybe there are higher numbers? Let me think.

Wait, but perhaps there are higher numbers where a>5? Because if a is higher, like 6,7,8,9, then the number would be larger. But according to our equations, a must be 5. Hmm, so that suggests that maybe 5694 is the maximum? But maybe I made a mistake in the equations?

Wait, let me check again the equations. Let me go back step by step.

We had equation7: a + c ≡0 mod7. So a = -c mod7. So if c is smaller, a can be larger? For example, if c=0, then a≡0 mod7, so a=7. But in our previous steps, we found that c must be 2 or 9. Wait, why?

Wait, equation8 came from equation3 - equation1: 6a -3c ≡3 mod7. Then, substituting a ≡-c mod7 into that equation led us to c ≡2 mod7. Wait, let me retrace that step.

Equation8 was 6a -3c ≡3 mod7. Then substituting a ≡-c mod7, so a = -c +7k for some integer k. Since a and c are digits, k is limited. Let me substitute a = -c mod7 into equation8:

6*(-c) -3c ≡3 mod7 →-6c -3c =-9c ≡3 mod7 →-9c ≡3 mod7 →Multiply both sides by -1: 9c ≡-3 mod7 →9 mod7 is 2, so 2c ≡-3 mod7 →2c ≡4 mod7 (since -3 mod7 is 4). Therefore, 2c ≡4 mod7 →c ≡2 mod7. So c must be congruent to 2 mod7, so possible c=2 or 9 (since 2+7=9, next is 16 which is too big). So that's correct. So c must be 2 or 9, so a must be 5. Therefore, a cannot be higher than 5? Hmm, so that would mean that the maximum possible a is 5. Therefore, the maximum N would be 5694. But is there a way to get a higher a?

Wait, perhaps my mistake is in the substitution steps. Let me check equation7 again. equation7 was from equation2 - equation1 substitution. Wait, let me retrace equation2 and equation1.

Wait, equation2 was 6a +3c +d ≡5 mod7.

Equation1 was 2b +3c +d ≡1 mod7.

Subtracting equation1 from equation2 gives 6a -2b ≡4 mod7. Wait, that was equation from equation2 - equation1. Wait, but then later, when I considered equation3 - equation1, I got equation8:6a -3c ≡3 mod7.

Wait, but perhaps there's another way to approach this. Let me see.

Alternatively, maybe there's an error in assuming that c must be exactly 2 or 9. Because when we have c ≡2 mod7, c can be 2 or 9, but perhaps when combined with other equations, there might be more possibilities? Wait, but c is a single digit, so only 2 and 9. So that's correct.

Alternatively, perhaps there is a higher a if we consider different congruencies? Let me see.

Wait, let me think differently. Since we need the largest possible N, which is four-digit, so starting from 9999 and going down, but maybe we can find a number higher than 5694 that satisfies the conditions. Let me try with a=9. Let's see if that's possible.

Suppose a=9. Then, from equation7: a +c ≡0 mod7 →9 +c ≡0 mod7 →c≡-9 mod7≡-2 mod7≡5 mod7. So c=5 or 12, but c is a digit so c=5 or 12 invalid. So c=5.

So if a=9, then c=5. Let's see if that works.

Then, with a=9, c=5, let's see other equations.

Equation5: d ≡3c -2 mod7 →3*5 -2=15-2=13 mod7 →13 mod7=6. So d≡6 mod7. So d can be 6 or 13, but 13 is invalid, so d=6.

Equation6: 2b +6c ≡3 mod7. c=5, so 6c=30 mod7. 30 mod7: 28 is 4*7, so 30 mod7=2. So equation6: 2b +2 ≡3 mod7 →2b ≡1 mod7. So 2b≡1 mod7. Multiply both sides by inverse of 2 mod7, which is 4. So b≡4 mod7. So b=4 or 11, but b is a digit, so b=4 or 11 invalid. So b=4.

So then, the number would be a=9, b=4, c=5, d=6 →9456. Let's check if this satisfies all four conditions.

Check each digit changed to1:

1. Change thousands digit to1: 1456. Check divisibility by7. 1456 ÷7. 7*200=1400, 1456-1400=56. 56 ÷7=8. So total 208. 7*208=1456. Okay.

2. Change hundreds digit to1: 9156. 9156 ÷7. Let's see 7*1300=9100, 9156-9100=56. 56 ÷7=8. So 1308. 7*1308=9156. Correct.

3. Change tens digit to1: 9416. 9416 ÷7. Let's compute 7*1345=9415, so 9416-9415=1. Not divisible. Wait, that's a problem. Hmm, so 9416 divided by7 gives remainder 1. So that violates the condition. Therefore, this number doesn't work. So something's wrong here.

Wait, so according to our equations, this should work, but in reality it doesn't. So perhaps my equations are missing something?

Wait, let me check equation3 for this case. equation3 was 6a +2b +d ≡4 mod7. Let's plug in a=9, b=4, d=6:

6*9 +2*4 +6 =54 +8 +6=68. 68 mod7: 7*9=63, 68-63=5 mod7. But equation3 requires it to be ≡4 mod7. 5≡4 mod7? No, so that's a contradiction. So my mistake must be in the equations or substitutions.

Wait, so where did I go wrong?

Wait, let's recast all equations for a=9, c=5, b=4, d=6.

Equation1: 2b +3c +d ≡2*4 +3*5 +6=8+15+6=29 mod7. 29 mod7: 28 is 4*7, so 29 mod7=1. Which matches equation1's requirement of ≡1 mod7. Okay.

Equation2:6a +3c +d=6*9 +15 +6=54+15+6=75 mod7. 75 mod7: 7*10=70, 75-70=5 mod7. Which matches equation2's requirement of ≡5 mod7. Okay.

Equation3:6a +2b +d=54 +8 +6=68 mod7=5 mod7. But equation3 requires ≡4 mod7. So that's a problem. So this combination fails equation3. So why did our earlier steps not catch this?

Wait, in equation6, which was derived from equation1 and equation5, we had equation6:2b +6c ≡3 mod7. Let's check with a=9, c=5, b=4:

2*4 +6*5=8+30=38 mod7. 38 mod7: 35 is 5*7, so 3 mod7. 38 mod7=3. Which matches equation6's requirement of ≡3 mod7. So that's okay.

Equation4:6a +2b +3c ≡6*9 +8 +15=54+8+15=77 mod7. 77 mod7=0. But equation4 requires ≡6 mod7. Wait, 77 mod7 is 0, which is not 6 mod7. So that's another problem. So equation4 is not satisfied.

Hmm, so this combination fails both equation3 and equation4. So my mistake must have been in the derivation steps. Let me check equation4 again.

Equation4 was 6a +2b +3c ≡6 mod7. For a=9, b=4, c=5: 54 +8 +15=77≡0 mod7≡6 mod7? No. So that's not okay.

Wait, so where did I go wrong in the equations?

Wait, let me retrace equation4's derivation. Equation4 came from the fourth condition: changing the units digit to1. The original number is N = a*1000 + b*100 +c*10 +d. Changing units digit to1 gives N' = a*1000 +b*100 +c*10 +1. The congruence is N' ≡0 mod7.

Expressed mod7, N' ≡6a +2b +3c +1 mod7 ≡0 mod7. So 6a +2b +3c ≡-1 mod7 ≡6 mod7. So equation4 is correct.

So in this case, with a=9, b=4, c=5, equation4 gives 6*9 +2*4 +3*5 =54 +8 +15=77≡0 mod7, which is not 6 mod7. So that's a problem. So why did our earlier steps allow this?

Wait, in the equations, when we set a=9 and c=5 (since a +c ≡0 mod7: 9+5=14≡0 mod7), then equation5: d≡3c -2 mod7 →3*5 -2=13≡6 mod7, so d=6. Then equation6:2b +6c ≡3 mod7 →2b +30≡3 mod7 →2b +2≡3 mod7 (since 30 mod7 is 2) →2b≡1 mod7 →b=4 mod7 (since 2*4=8≡1 mod7). So b=4.

So according to equations 1-5, this should satisfy all, but equations3 and4 are not satisfied. So there must be an error in the derivation steps. Let me check equation3 again.

Equation3 was derived from the third condition (changing tens digit to1):

The number becomes a*1000 +b*100 +1*10 +d. Its mod7 is 6a +2b +3 +d ≡0 mod7. So equation3:6a +2b +d ≡-3 mod7≡4 mod7.

So with a=9, b=4, d=6: 6*9 +2*4 +6 =54 +8 +6=68 mod7. 68 divided by7 is 9*7=63, remainder5. So 5 mod7, which is not 4 mod7. So discrepancy here.

Wait, so where is the mistake?

Wait, perhaps I made a mistake in the substitution steps when I subtracted equations. Let me check equation8 again.

Equation8 was from equation3 - equation1:

Equation3:6a +2b +d ≡4 mod7.

Equation1:2b +3c +d ≡1 mod7.

Subtract equation1 from equation3: (6a +2b +d) - (2b +3c +d) =6a -3c ≡4 -1 mod7 →6a -3c ≡3 mod7. So equation8 is correct.

Then, with a=9 and c=5, 6a -3c =54 -15=39 mod7. 39 mod7: 35 is 5*7, so 4 mod7. 39 mod7=4. But equation8 requires 6a -3c ≡3 mod7. 4≡3 mod7? No. So that's a problem. Wait, so this combination doesn't satisfy equation8 either. But according to equation7 and equation8, when a=9 and c=5, equation7 is satisfied (a +c=14≡0 mod7), but equation8 is not. So that means that even with a=9 and c=5, which satisfy equation7, they don't satisfy equation8. Therefore, my earlier assumption that c must be 2 mod7 is still valid, but when I tried to set a=9 and c=5 (which is 5≡-9 mod7?), let me check:

Wait, a +c ≡0 mod7. If a=9, then c must be -9 mod7≡-2 mod7≡5 mod7. So c=5 mod7. So c can be 5 or 12 (invalid). So c=5. But then equation8 requires 6a -3c ≡3 mod7. Plugging a=9, c=5: 6*9 -3*5=54-15=39≡4 mod7≡3 mod7? No. So that combination doesn't work. Therefore, even though equation7 is satisfied, equation8 is not. Therefore, there is no solution with a=9. So my mistake was in thinking that setting a=9 and c=5 would work, but it doesn't satisfy equation8. Therefore, such a number with a=9 doesn't exist. So the earlier conclusion that c must be 2 mod7 (so c=2 or 9) and a=5 is correct, and thus the maximum a is 5. Therefore, the maximum N is 5694.

Wait, but let me check with a=8. Maybe a=8 can work?

Let me try a=8. Then equation7: a +c ≡0 mod7 →8 +c ≡0 mod7 →c≡-8 mod7≡-1 mod7≡6 mod7. So c=6 or 13 (invalid). So c=6.

Then, with a=8, c=6:

Equation5: d≡3c -2 mod7 →3*6 -2=16 mod7 →16 mod7=2. So d≡2 mod7. So d=2 or 9.

Equation6:2b +6c ≡3 mod7. 6c=36 mod7. 36 mod7=1 (since 35 is 5*7). So equation6:2b +1 ≡3 mod7 →2b≡2 mod7 →b≡1 mod7. So b=1 or 8.

So possible b=1 or 8.

Let me consider b=8 (to maximize the number). So b=8, then the number would be a=8, b=8, c=6, d=2 or 9. Let's choose d=9 (larger).

Wait, d must be ≡2 mod7, so d can be 2 or 9 (since 2 mod7 and 9 mod7 is 2). So d=9 is allowed. So let's take d=9. So the number is 8869.

Check equations:

First, check equation3 and equation4.

Equation3:6a +2b +d ≡6*8 +2*8 +9=48+16+9=73 mod7. 73 mod7: 70 is 10*7, so 3 mod7. Equation3 requires ≡4 mod7. Not okay. Hmm.

Alternatively, let's see step by step.

Wait, let's first check all equations for a=8, c=6, b=8, d=9.

Equation1:2b +3c +d=16 +18 +9=43 mod7. 43 mod7: 42 is 6*7, so 1 mod7. Which matches equation1's 1 mod7. Good.

Equation2:6a +3c +d=48 +18 +9=75 mod7=5 mod7 (since 70 is 10*7, 75-70=5). Which matches equation2's 5 mod7. Good.

Equation3:6a +2b +d=48 +16 +9=73 mod7=73-70=3 mod7. But needs to be 4 mod7. Not good.

Equation4:6a +2b +3c=48 +16 +18=82 mod7. 82 mod7: 77 is 11*7, so 5 mod7. Equation4 requires 6 mod7. Not okay.

Hmm, so that doesn't work. Let's try with b=1 instead of 8. So b=1, d=9 (to maximize):

Number is 8 1 6 9 →8169.

Check equations:

Equation3:6*8 +2*1 +9=48+2+9=59 mod7. 59 mod7: 56 is 8*7, so 3 mod7. Still not 4 mod7.

Equation4:6*8 +2*1 +3*6=48+2+18=68 mod7=5 mod7. Not 6 mod7.

Alternatively, d=2 instead of 9. Let's try d=2.

With a=8, c=6, b=8, d=2: number 8862.

Equation3:6*8 +2*8 +2=48+16+2=66 mod7. 66 mod7: 63 is 9*7, so 3 mod7. Still not 4.

Equation4:6*8 +2*8 +3*6=48+16+18=82 mod7=5 mod7. Not okay.

Hmm, so even with a=8, it's not working. Maybe b=1 and d=2: 8162.

Equation3:6*8 +2*1 +2=48+2+2=52 mod7. 52 mod7: 49 is 7*7, so 3 mod7. Still no.

Hmm, so perhaps a=8 is not possible. Let me try with a=7.

a=7. Then equation7: a +c ≡0 mod7 →7 +c ≡0 mod7 →c≡0 mod7. So c=0 or7.

Case1: c=0.

Then equation5: d≡3c -2 mod7 →0 -2 ≡-2 mod7≡5 mod7. So d=5 or 12→5.

Equation6:2b +6c ≡3 mod7 →2b +0 ≡3 mod7 →2b≡3 mod7 →b≡5 mod7 (since 2*5=10≡3 mod7). So b=5 or 12→5.

Thus, the number would be a=7, b=5, c=0, d=5 →7505.

Check equations:

Equation3:6*7 +2*5 +5=42+10+5=57 mod7. 57 mod7: 49 is 7*7, so 8 mod7=1 mod7. Not 4 mod7. Not good.

Alternatively, let's check all conditions:

Changing thousands digit to1:1505. 1505 ÷7. 7*215=1505. Yes, divisible.

Changing hundreds digit to1:7105. 7105 ÷7. 7*1000=7000, 7105-7000=105. 105 ÷7=15. So yes.

Changing tens digit to1:7515. 7515 ÷7. 7*1000=7000, 7515-7000=515. 515 ÷7≈73.57. Not divisible. 7*73=511, 515-511=4. So remainder 4. Not divisible. So this fails.

Hmm, so even though equations1-2 are satisfied, equation3 fails. So this approach isn't working.

Alternatively, with c=7 (since c≡0 mod7, so c=0 or7). Let's try c=7.

a=7, c=7.

Equation5: d≡3*7 -2=21-2=19 mod7 →19 mod7=5. So d=5 or 12→5.

Equation6:2b +6*7 ≡3 mod7 →2b +42 mod7. 42 mod7=0, so 2b≡3 mod7 →b≡5 mod7. So b=5 or 12→5.

Thus, number is 7 5 7 5 →7575.

Check equation3:6*7 +2*5 +5=42+10+5=57 mod7≡1 mod7≠4 mod7. Not good.

Changing tens digit to1:7515, same as before, which isn't divisible by7. So no.

Hmm, so a=7 also not working.

Therefore, perhaps the only valid solutions are when a=5, c=2 or9, leading to 5624 and 5694. Since 5694 is larger, that's the candidate. Let me check if there are other possibilities with a=5 but higher digits.

Wait, in the case of c=9, we had a=5, b=6, c=9, d=4. So 5694. If we can make b higher than6, that would be better. Let me see.

Wait, in the case of c=9, we had equation6:2b +6c ≡3 mod7. With c=9, 6c=54 mod7=54-49=5 mod7. So equation6:2b +5 ≡3 mod7 →2b≡-2 mod7→2b≡5 mod7→b≡6 mod7. So b must be6 or 13, but since b is a digit, only 6. So b can't be higher. So 5694 is the maximum in that case.

Alternatively, if c=2, then b=6 as well, so 5624. So 5694 is the larger one.

Wait, but let me check if there are other values for c beyond 2 and9. Wait, c must be ≡2 mod7, so only 2 and9. So no.

Alternatively, maybe there are other solutions where the congruences are satisfied with different a, but I don't think so because of the way the equations are set up.

Therefore, the maximum N is 5694. Let me confirm once again that changing any digit to1 gives a multiple of7:

1. Change thousands digit to1:1694. 1694 ÷7=242. 7*242=1694. Good.

2. Change hundreds digit to1:5194. 5194 ÷7=742. 7*742=5194. Good.

3. Change tens digit to1:5614. 5614 ÷7=802. 7*802=5614. Good.

4. Change units digit to1:5691. 5691 ÷7=813. 7*813=5691. Good.

All conditions are satisfied. So N=5694.

Now, compute Q and R when N is divided by1000.

N=5694. Divided by1000, quotient Q=5, remainder R=694. Therefore Q+R=5+694=699.

Wait, but let me confirm:

5694 divided by1000 is 5.694, so quotient is 5, remainder is 5694 -5*1000=694. So yes, Q=5, R=694. So Q+R=5+694=699.

But before I conclude, let me check if there is a larger N than 5694 that satisfies the conditions. For example, maybe 5774 or something? Let me try with a=5, but higher b or c.

Wait, let's see. Suppose a=5, and c=9, which gives the highest possible number. Since c=9 is the maximum possible c (as c can be 2 or9). Then, with c=9, b must be6, d must be4. So 5694 is the maximum in that case. If we try to increase b beyond6, but equation6 requires b≡6 mod7, so next would be 13, which is invalid. So can't.

Alternatively, if there's another solution with a=5, c=9, but different b and d? Let me see.

Wait, in equation5, d must be ≡3c -2 mod7. With c=9, that's 25 mod7=4 mod7, so d must be4 or11 (invalid). So d=4. Equation6 requires b=6. So no other options.

Alternatively, maybe there's another set of equations where c is higher but with different congruence? But according to our earlier analysis, no. So 5694 is indeed the maximum.

Therefore, the answer is Q+R=5+694=699.

Wait, but let me check another number like 5694 +7=5701. But changing a digit to1 might not work. Let me see 5701. But it's not necessary, since our analysis shows 5694 is the maximum.

Alternatively, perhaps there is a number between 5694 and 9999 that satisfies the conditions. Let me think of 5774. Let me check.

Wait, let me try N=5774. Let's see if it meets the conditions.

Changing thousands digit to1:1774. 1774 ÷7. 7*253=1771, 1774-1771=3. Not divisible.

So no. How about 5764?

Changing thousands digit to1:1764. 1764 ÷7=252. Okay.

Changing hundreds digit to1:5164. 5164 ÷7. 7*700=4900, 5164-4900=264. 264 ÷7≈37.7. Not divisible. So no.

Hmm, so probably 5694 is the correct answer. Therefore, I think the answer is 699.

**Final Answer**
\boxed{699}
</think>

Given a four-digit number \( N = 1000a + 100b + 10c + d \), we need to ensure that changing any single digit to 1 results in a number divisible by 7. We derived the following congruences modulo 7:

1. Changing the thousands digit to 1: \( 6 + 2b + 3c + d \equiv 0 \pmod{7} \)
2. Changing the hundreds digit to 1: \( 6a + 2 + 3c + d \equiv 0 \pmod{7} \)
3. Changing the tens digit to 1: \( 6a + 2b + 3 + d \equiv 0 \pmod{7} \)
4. Changing the units digit to 1: \( 6a + 2b + 3c + 1 \equiv 0 \pmod{7} \)

Through solving these congruences, we found that \( a = 5 \), \( c \) must be either 2 or 9, and \( b \) and \( d \) are determined accordingly. The largest valid number is when \( c = 9 \), leading to \( b = 6 \) and \( d = 4 \), forming the number 5694.

Verifying 5694:
1. Changing thousands digit to 1: 1694 is divisible by 7.
2. Changing hundreds digit to 1: 5194 is divisible by 7.
3. Changing tens digit to 1: 5614 is divisible by 7.
4. Changing units digit to 1: 5691 is divisible by 7.

Thus, \( N = 5694 \). Dividing by 1000 gives quotient \( Q = 5 \) and remainder \( R = 694 \). Therefore, \( Q + R = 5 + 694 = 699 \).

\[
\boxed{699}
\]