Difference between revisions of "2010 AMC 10A Problems/Problem 24"
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<math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math> | <math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math> | ||
− | == Solution == | + | == Solution 1(Bigbrain) == |
We will use the fact that for any integer <math>n</math>, <cmath> | We will use the fact that for any integer <math>n</math>, <cmath> | ||
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Finally, combining with the fact that <math>N\equiv 0\pmod 4</math> yields <math>n=\boxed{\textbf{(A)}\ 12}</math>. | Finally, combining with the fact that <math>N\equiv 0\pmod 4</math> yields <math>n=\boxed{\textbf{(A)}\ 12}</math>. | ||
+ | == Solution 2(bash) == | ||
== See also == | == See also == |
Revision as of 23:49, 28 January 2020
Problem
The number obtained from the last two nonzero digits of is equal to . What is ?
Solution 1(Bigbrain)
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 . Since there is clearly an excess of factors of 2, we know that , so it remains to find .
We can write as where where every number in the form is replaced by .
The number can be grouped as follows:
Hence, we can reduce to
Using the fact that ,we can deduce that . Therefore .
Finally, combining with the fact that yields .
Solution 2(bash)
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
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