Difference between revisions of "2000 AIME I Problems/Problem 1"
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== Solution == | == Solution == | ||
− | If a factor of <math>10^{n}</math> has a <math>2</math> and a <math>5</math> in its [[prime factorization]], then that factor will end in a <math>0</math>. Therefore, we have left to consider the case when the two factors have the <math>2</math>s and the <math>5</math>s separated, | + | If a factor of <math>10^{n}</math> has a <math>2</math> and a <math>5</math> in its [[prime factorization]], then that factor will end in a <math>0</math>. Therefore, we have left to consider the case when the two factors have the <math>2</math>s and the <math>5</math>s separated, so we need to find the first power of 2 or 5 that contains a 0. |
− | + | For <math>n = 1:</math> <cmath>2^1 = 2 , 5^1 = 5</cmath> | |
− | + | <math>n = 2:</math> <cmath>2^2 = 4 , 5 ^ 2 =25</cmath> | |
− | + | <math>n = 3:</math> <cmath>2^3 = 8 , 5 ^3 = 125</cmath> | |
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− | We see that <math>5^8</math> | + | and so on, until, |
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+ | <math>n = 8:</math> <math>2^8 = 256</math> | <math>5^8 = 390625</math> | ||
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+ | We see that <math>5^8</math> contains the first zero, so <math>n = \boxed{8}</math>. | ||
== See also == | == See also == | ||
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:44, 16 December 2020
Problem
Find the least positive integer such that no matter how
is expressed as the product of any two positive integers, at least one of these two integers contains the digit
.
Solution
If a factor of has a
and a
in its prime factorization, then that factor will end in a
. Therefore, we have left to consider the case when the two factors have the
s and the
s separated, so we need to find the first power of 2 or 5 that contains a 0.
For
and so on, until,
|
We see that contains the first zero, so
.
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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.