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.
Solutions
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 2
Let be the required number, and be with the first digit deleted. Now, we know that (because this is an AIME problem). Thus, has or digits. Checking the other cases, we see that it must have digits. Let , so . Thus, . By the constraints of the problem, we see that , so Now, we subtract and divide to get Clearly, must be a multiple of because both and are multiples of . Thus, . Now, we plug that into the equation: By the same line of reasoning as earlier, . We again plug that into the equation to get Now, since , , and , our number .
Here's another way to finish using this solution. From the above, you have Divide by , and you get This means that has to be divisible by , and hence Now, solve for , which gives you , giving you the number
Solution 3 (Quick)
Note that if we let the last digit be we must have Thus we either have which we can quickly check to be impossible (since the number after digit removal could be 10,20,30) or Testing 5, 15, and 25 as the numbers after removal we find that our answer is clearly
~Dhillonr25
(Note that the quick checking of six numbers was possible thanks to AIME problems having answers less than 1000).
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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