Difference between revisions of "1994 AIME Problems/Problem 5"
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== 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> | + | 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>2115\cdot (1+1+2+3+4+5+6+7+8+9)=2115\cdot 46=47\cdot 45\cdot 46</math>. But we didn't count 100, 200, 300, ..., 900. We add another 45 to get <math>45\cdot 2163</math>. The largest prime factor of that is <math>\boxed{103}</math>. |
== See also == | == See also == | ||
Revision as of 14:22, 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
.
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 
. But we didn't count 100, 200, 300, ..., 900. We add another 45 to get 
. 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 | ||
