Difference between revisions of "1968 AHSME Problems/Problem 21"
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== Solution == | == Solution == | ||
| − | 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! \equiv 3 mod 10</math>, which means that the unit digit is <math>3</math>, we have our answer of <math>\fbox{D}</math> as desired. | + | 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! \equiv 3</math> mod <math>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 == | ||
| − | {{AHSME box|year=1968|num-b=20|num-a=22}} | + | {{AHSME 35p box|year=1968|num-b=20|num-a=22}} |
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:53, 16 August 2023
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 
mod 
, which means that the unit digit is 
, we have our answer of 
as desired.
See also
| 1968 AHSC (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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
