Difference between revisions of "2005 AMC 10A Problems/Problem 22"
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Since <math>4\cdot3=12</math> every <math>3</math>rd element of <math>S</math> will be a multiple of <math>12</math> | Since <math>4\cdot3=12</math> every <math>3</math>rd element of <math>S</math> will be a multiple of <math>12</math> | ||
− | Therefore the answer is <math>\lfloor\frac{2005}{3}\rfloor=668\Longrightarrow \mathrm{(D)}</math> | + | Therefore the answer is <math>\lfloor\frac{2005}{3}\rfloor=668\Longrightarrow \boxed{\mathrm{(D) 668}}</math> |
==See Also== | ==See Also== |
Revision as of 20:03, 8 August 2017
Problem
Let be the set of the smallest positive multiples of , and let be the set of the smallest positive multiples of . How many elements are common to and ?
Solution
Since the least common multiple , the elements that are common to and must be multiples of .
Since and , several multiples of that are in won't be in , but all multiples of that are in will be in . So we just need to find the number of multiples of that are in .
Since every rd element of will be a multiple of
Therefore the answer is
See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AMC 10 Problems and Solutions |
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