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# 2006 AIME I #3

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+736

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.

Jun 27, 2021

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let the resulting integer equal x. reword it another way, $$29x\equiv x(\text{mod }10^n)\rightarrow28x\equiv 0(\text{mod }10^n)$$where n is just some positive integer.

the prime factorization of 28 is 2*2*7, meaning that there must be at least a 5 in the prime factorization of x to multiply with one of the 2's in 28(for its last digit to equal zero and for n to be 1). If we were to plug 5 in, we see that it doesn't work, meaning that x must be at least 2 digits. That means that x must have at least 2 5's in its prime factorization for it to multiply with the 2 2's in 28(for its last 2 digits to equal zero and for n to be 2). Testing out 5*5=25, we can see that it works, and we are sure it's the smallest possible value, so the answer is $$25*29=\boxed{725}$$

Jun 27, 2021
edited by textot  Jun 27, 2021
edited by textot  Jun 27, 2021