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 | + | 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>. | <center><math>S=p(1)+p(2)+p(3)+\cdots+p(999)</math></center>. | ||
What is the largest prime factor of <math>S\,</math>? | 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}} | {{AIME box|year=1994|num-b=4|num-a=6}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:27, 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
![$S=p(1)+p(2)+p(3)+\cdots+p(999)$](http://latex.artofproblemsolving.com/3/4/d/34d0f62dc5536d2351b2e6ecbf4226fcabf7ab4d.png)
.
What is the largest prime factor of ?
Solution
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
![$(1+ 1 +2+3+4+5+6+7+8+9) (1+ 1 +2+3+\cdots+9) (1+ 1 +2+3+\cdots+9)$](http://latex.artofproblemsolving.com/7/0/8/708bf6bfc4a67a6c8ccf6ab98d670c43075981ef.png)
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 |
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