Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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\...)
 
(category)
 
(One intermediate revision by one other user not shown)
Line 8: Line 8:
 
==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 20:45, 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
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_6&oldid=22431"