Difference between revisions of "2024 AIME II Problems/Problem 7"

Revision as of 20:39, 9 February 2024

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

Solution 3

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

Video Solution

https://youtu.be/DxBjaFJneBs

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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 {{AIME box|year=2024|num-b=6|num-a=8|n=II}}
-[[Category:]]
+[[Category:Intermediate Number Theory Problems]]
 {{MAA Notice}}

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