Difference between revisions of "1968 AHSME Problems/Problem 21"

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== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
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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 ==

Latest revision as of 07:36, 11 July 2019

Problem

If $S=1!+2!+3!+\cdots +99!$, then the units' digit in the value of S is:

$\text{(A) } 9\quad \text{(B) } 8\quad \text{(C) } 5\quad \text{(D) } 3\quad \text{(E) } 0$

Solution

Note that every factorial after $5!$ has a unit digit of $0$, meaning that we can disregard them. Thus, we only need to find the units digit of $1! + 2! + 3! +  4!$, and as $1! + 2! + 3! +  4! \equiv 3$ mod $10$, which means that the unit digit is $3$, we have our answer of $\fbox{D}$ as desired.

See also

1968 AHSME (ProblemsAnswer KeyResources)
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

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