Difference between revisions of "2019 AMC 10A Problems/Problem 2"
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− | 20 and 15 are both greater than 10, therefore they are divisible by 100 because of powers of 5 and powers of 2, so the hundreds digit is <math>\boxed{\textbf{(A) }0}</math>. ~ | + | 20 and 15 are both greater than 10, therefore they are divisible by 100 because of powers of 5 and powers of 2, so the hundreds digit is <math>\boxed{\textbf{(A) }0}</math>. ~peppapig_ |
==See Also== | ==See Also== |
Revision as of 20:59, 1 April 2020
Contents
Problem
What is the hundreds digit of
Solution
The last three digits of for all are , because there are at least three s and three s in its prime factorization. Because , the answer is .
Solution 2
20 and 15 are both greater than 10, therefore they are divisible by 100 because of powers of 5 and powers of 2, so the hundreds digit is . ~peppapig_
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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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