2024 AIME II Problems/Problem 7

Problem

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 $Q+R$ .

Solution 1

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

Solution 2

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

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2024 AIME II Problems/Problem 7

Problem

Solution 1

Solution 2