2010 AMC 12A Problems/Problem 23
Contents
[hide]Problem
The number obtained from the last two nonzero digits of is equal to . What is ?
Solution
We will use the fact that for any integer ,
First, we find that the number of factors of in is equal to . Let . The we want is therefore the last two digits of , or . If instead we find , we know that , what we are looking for, could be , , , or . Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, because has to be a multiple of 4.
If we divide by by taking out all the factors of in , we can write as where where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .
The number can be grouped as follows:
Where the first line is composed of the numbers in that aren't multiples of five, the second line is the multiples of five and not 25 after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.
Using the identity at the beginning of the solution, we can reduce to
Using the fact that (or simply the fact that if you have your powers of 2 memorized), we can deduce that . Therefore .
Finally, combining with the fact that yields .
Solution 2
Let be after we truncate its zeros. We shall consider modulo 4 and 25, to determine its residue modulo 100. It is easy to prove that is divisible by 4 (consider the number of 2s dividing minus the number of 5s dividing ), and so we only need to consider modulo 25.
Now, notice that for integers we have
Thus, for integral a:
Notice that differs from by an exponential of 10. Thus, we can "divide out" all the zeros from 10, 20, 30, ..., 90. We are left with the integers from 1 to 9, from which we multiply the non-five numbers to obtain a . Now, we combine this 5 with the remaining 5, 15, 25, ..., 85. To divide by 10 is to cancel a 5 from a number and divide by 2. Because 25 is relatively prime to 2, we can transform the "divide by 2" into multiply by modulo 25. Thus, the 5, 5, 15, 25, ..., 85 becomes 1, 1, 3, 1, 7, 9, 11, 13, 3, 17 and . We deduce that from multiplying out the 1, 1, 3, 1, 7, ..., 17 is equivalent to 2 modulo 25, and so we need to compute . But this is simply by Fermat's Little Theorem . Because 12 is also a multiple of 4, we can utilize the Chinese Remainder Theorem to show that and so the answer is .
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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