Difference between revisions of "1994 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
| + | Given a positive integer <math>n\,</math>, let <math>p(n)\,</math> be the product of the non-zero digits of <math>n\,</math>. (If <math>n\,</math> has only one digit, then <math>p(n)\,</math> is equal to that digit.) Let | ||
| + | <center><math>S=p(1)+p(2)+p(3)+\cdots+p(999)</math></center>. | ||
| + | What is the largest prime factor of <math>S\,</math>? | ||
| + | |||
| + | __TOC__ | ||
== Solution == | == Solution == | ||
| + | === Solution 1 === | ||
| + | Suppose we write each number in the form of a three-digit number (so <math>5 \equiv 005</math>), and since our <math>p(n)</math> ignores all of the zero-digits, replace all of the <math>0</math>s with <math>1</math>s. Now note that in the expansion of | ||
| + | |||
| + | <center><math>(1+ 1 +2+3+4+5+6+7+8+9) (1+ 1 +2+3+\cdots+9) (1+ 1 +2+3+\cdots+9)</math></center> | ||
| + | |||
| + | we cover every permutation of every product of <math>3</math> digits, including the case where that first <math>1</math> represents the replaced <math>0</math>s. However, since our list does not include <math>000</math>, we have to subtract <math>1</math>. Thus, our answer is the largest prime factor of <math>(1+1+2+3+\cdots +9)^3 - 1 = 46^3 - 1 = (46-1)(46^2 + 46 + 1) = 3^3 \cdot 5 \cdot 7 \cdot \boxed{103}</math>. | ||
| + | |||
| + | === Solution 2 === | ||
| + | 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 == | ||
| − | + | {{AIME box|year=1994|num-b=4|num-a=6}} | |
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:27, 14 June 2024
Problem
Given a positive integer 
, let 
be the product of the non-zero digits of . (If has only one digit, then is equal to that digit.) Let
.
What is the largest prime factor of 
?
Solution
Solution 1
Suppose we write each number in the form of a three-digit number (so 
), and since our 
ignores all of the zero-digits, replace all of the 
s with 
s. Now note that in the expansion of
we cover every permutation of every product of 
digits, including the case where that first represents the replaced s. However, since our list does not include 
, we have to subtract . Thus, our answer is the largest prime factor of 
.
Solution 2
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
