Difference between revisions of "2007 iTest Problems/Problem 6"

(New page: ==Problem== Find the units digit of the sum <cmath>\sum_{i=1}^{100}(i!)^{2}</cmath> <math>\mathrm{(A)}\,0\quad\mathrm{(B)}\,1\quad\mathrm{(C)}\,3\quad\mathrm{(D)}\,5\quad\mathrm{(E)}\,7\...)
 
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==Solution==
 
==Solution==
 
If i is less than 5, then <math>i!</math> has a positive units digit, if <math>i\geq 5</math>, then <math>i!</math> has a units digit of 0, as does <math>(i!)^2</math>. So we only need to worry about i=1-4.
 
If i is less than 5, then <math>i!</math> has a positive units digit, if <math>i\geq 5</math>, then <math>i!</math> has a units digit of 0, as does <math>(i!)^2</math>. So we only need to worry about i=1-4.
 +
*<math>(1!)^2=1</math>
 +
*<math>(2!)^2=4</math>
 +
*<math>(3!)^2=36</math>
 +
*<math>(4!)^2=576</math>
 +
*<math>1+4+6+6=17</math>, which has a units digit of 7 <math>\Rightarrow \mathrm{(E)}</math>
  
<math>(1!)^2=1</math>
+
==See Also==
 +
{{iTest box|year=2007|num-b=5|num-a=7}}
  
<math>(2!)^2=4</math>
+
[[Category:Introductory Number Theory Problems]]
 
 
<math>(3!)^2=36</math>
 
 
 
<math>(4!)^2=576</math>
 
 
 
<math>1+4+6+6=17</math>, which has a units digit of 7 <math>\Rightarrow \mathrm{(E)}</math>
 

Latest revision as of 19:45, 14 January 2008

Problem

Find the units digit of the sum

\[\sum_{i=1}^{100}(i!)^{2}\]

$\mathrm{(A)}\,0\quad\mathrm{(B)}\,1\quad\mathrm{(C)}\,3\quad\mathrm{(D)}\,5\quad\mathrm{(E)}\,7\quad\mathrm{(F)}\,9$

Solution

If i is less than 5, then $i!$ has a positive units digit, if $i\geq 5$, then $i!$ has a units digit of 0, as does $(i!)^2$. So we only need to worry about i=1-4.

  • $(1!)^2=1$
  • $(2!)^2=4$
  • $(3!)^2=36$
  • $(4!)^2=576$
  • $1+4+6+6=17$, which has a units digit of 7 $\Rightarrow \mathrm{(E)}$

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 5
Followed by:
Problem 7
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