Difference between revisions of "1987 AJHSME Problems/Problem 9"
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==Solution== | ==Solution== | ||
| − | We want the [[least common multiple]] of <math>2,3,4,5,6,7</math>, which is <math>420</math>, or choice <math>\boxed{\text{C}}</math>. | + | We want the [[least common multiple]] of <math>2,3,4,5,6,7</math>, which is <math>420</math>, or choice <math>\boxed{\text{C}}</math>. |
| + | |||
| + | ==Solution 2== | ||
| + | Since we want to find the LCM (least common multiple) of 2,3,4,5,6,7 we would be able to do this simply by finding the prime factorization of each of these numbers. First, we know that since 2 and 3 are already [[prime]] numbers, it would be completely factored. But, since 4 is a [[composite]] number, we would need to find the prime factorization of this, which would simply be 2 x 2. Afterward, we would basically just do this for all the numbers that are not prime (composite). Once we find the prime factorization of all of these numbers we would multiple the greatest prime factors. | ||
| + | |||
| + | 2 --> 2 | ||
| + | 3 --> 3 | ||
| + | 4 --> 2^2 | ||
| + | 5 --> 5 | ||
| + | 6 --> 2 x 3 | ||
| + | 7 --> 7 | ||
| + | |||
| + | Since 2^2 is the greatest number of 2's we can have we would simply multiply this by the greatest number of 3's, 5's, and 7's we could have in being the factors of each number. So, | ||
| + | |||
| + | 2^2 x 3 x 5 x 7 = 420 | ||
==See Also== | ==See Also== | ||
| − | {{AJHSME box|year= | + | {{AJHSME box|year=1987|num-b=8|num-a=10}} |
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:37, 29 December 2020
Contents
Problem
When finding the sum 
, the least common denominator used is
Solution
We want the least common multiple of 
, which is 
, or choice 
.
Solution 2
Since we want to find the LCM (least common multiple) of 2,3,4,5,6,7 we would be able to do this simply by finding the prime factorization of each of these numbers. First, we know that since 2 and 3 are already prime numbers, it would be completely factored. But, since 4 is a composite number, we would need to find the prime factorization of this, which would simply be 2 x 2. Afterward, we would basically just do this for all the numbers that are not prime (composite). Once we find the prime factorization of all of these numbers we would multiple the greatest prime factors.
2 --> 2 3 --> 3 4 --> 2^2 5 --> 5 6 --> 2 x 3 7 --> 7
Since 2^2 is the greatest number of 2's we can have we would simply multiply this by the greatest number of 3's, 5's, and 7's we could have in being the factors of each number. So,
2^2 x 3 x 5 x 7 = 420
See Also
| 1987 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
