Difference between revisions of "1968 AHSME Problems/Problem 21"
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== Solution == | == Solution == | ||
− | <math>\fbox{D}</math> | + | Note that every factorial after <math>5!</math> has a unit digit of <math>0</math>, meaning that we can disregard them. Thus, we only need to find the units digit of <math>1! + 2! + 3! + 4!</math>, and as <math>1! + 2! + 3! + 4! = 3 \equiv 10</math>, which means that the unit digit is <math>3</math>, we have our answer of <math>\fbox{D}</math> as desired. |
== See also == | == See also == |
Revision as of 15:03, 11 February 2017
Problem
If , then the units' digit in the value of S is:
Solution
Note that every factorial after has a unit digit of , meaning that we can disregard them. Thus, we only need to find the units digit of , and as , which means that the unit digit is , we have our answer of as desired.
See also
1968 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.