Difference between revisions of "1994 AIME Problems/Problem 5"
m |
|||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | { | + | Note that <math>p(1)=p(11), p(2)=p(12), p(3)=p(13), \cdots p(19)=p(9)</math>, and <math>p(37)=3p(7)</math>. So <math>p(10)+p(11)+p(12)+\cdots +p(19)=46</math>, <math>p(10)+p(11)+\cdots +p(99)=46*45=2070</math>. We add <math>p(1)+p(2)+p(3)+\cdots +p(10)=45</math> to get 2115. When we add a digit we multiply the sum by that digit. Thus <math>S=2115\cdot (1+1+2+3+4+5+6+7+8+9)=2115\cdot 46=47\cdot 45\cdot 46</math>. The largest prime factor of that is <math>\boxed{47}</math>. |
== See also == | == See also == | ||
{{AIME box|year=1994|num-b=4|num-a=6}} | {{AIME box|year=1994|num-b=4|num-a=6}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 14:17, 6 May 2008
Problem
Given a positive integer , let
be the product of the non-zero digits of
. (If
has only one digits, then
is equal to that digit.) Let
![$S=p(1)+p(2)+p(3)+\cdots+p(999)$](http://latex.artofproblemsolving.com/3/4/d/34d0f62dc5536d2351b2e6ecbf4226fcabf7ab4d.png)
.
What is the largest prime factor of ?
Solution
Note that , and
. So
,
. We add
to get 2115. When we add a digit we multiply the sum by that digit. Thus
. The largest prime factor of that is
.
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |