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

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2024 AIME II Problems/Problem 7

Problem

Solution 1