2024 AIME II Problems/Problem 7
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[hide]Problem
Let be the greatest four-digit positive integer with the property that whenever one of its digits is changed to , the resulting number is divisible by . Let and be the quotient and remainder, respectively, when is divided by . Find .
Solution 1
We note that by changing a digit to for the number , we are subtracting the number by either , , , or . Thus, . We can casework on backwards, finding the maximum value.
(Note that computing greatly simplifies computation).
Applying casework on , we can eventually obtain a working value of . ~akliu
Solution 2
Let our four digit number be . Replacing digits with 1, we get the following equations:
Reducing, we get
Subtracting , we get:
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 -westwoodmonster
Solution 3
Let our four digit number be . Replacing digits with , we get the following equations:
Add the equations together, we get:
And since the remainder of 1111 divided by 7 is 5, we get:
Which gives us:
And since we know that changing each digit into will make divisible by , we get that , , , and all have a remainder of when divided by . Thus, we get , , , and . Thus, we get as , and the answer is .
~Callisto531
Solution 4
Let our four digit number be . Replacing digits with 1, we get the following equations:
Then, we let x, y, z, t be the smallest whole number satisfying the following equations:
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):
(2):
(3):
(4):
Add (1), (2), (3) together, we get:
We can transform this equation to:
Since, according to (4), has a remainder of 0 when divided by 7, we get:
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
~Callisto531 and his dad
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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