2006 AIME I Problems/Problem 3
Problem
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is of the original integer.
Solution
Solution 1
Suppose the original number is where the are digits and the first digit, is nonzero. Then the number we create is so But is with the digit added to the left, so Thus, The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number is never divisible by so must be divisible by But is a nonzero digit, so the only possibility is This gives or Now, we want to minimize both and so we take and Then and indeed,
Solution 1
Suppose the original number is where the are digits and the first digit, is nonzero. Then the number we create is so But is with the digit added to the left, so Thus, The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number is never divisible by so must be divisible by But is a nonzero digit, so the only possibility is This gives or Now, we want to minimize both and so we take and Then and indeed,
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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