Difference between revisions of "2002 AMC 10P Problems/Problem 22"
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| + | == Problem == | ||
| + | |||
| + | In how many zeroes does the number <math>\frac{2002!}{(1001!)^2}</math> end? | ||
| + | |||
| + | <math> | ||
| + | \text{(A) }0 | ||
| + | \qquad | ||
| + | \text{(B) }1 | ||
| + | \qquad | ||
| + | \text{(C) }2 | ||
| + | \qquad | ||
| + | \text{(D) }200 | ||
| + | \qquad | ||
| + | \text{(E) }400 | ||
| + | </math> | ||
| + | |||
== Solution 1== | == Solution 1== | ||
Revision as of 19:01, 14 July 2024
Problem
In how many zeroes does the number 
end?
Solution 1
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
