Difference between revisions of "1981 AHSME Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | The least common multiple of 2, 3, 4, and 5 is <math>2^2 \cdot 3 \cdot 5 = 60 | + | The least common multiple of <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> is <math>2^2 \cdot 3 \cdot 5 = 60</math>. There is only one multiple of <math>60</math> between <math>1</math> and <math>100</math>, so the answer is <math>\textbf{(B)}\ 1.</math> |
-edited by coolmath34 | -edited by coolmath34 | ||
| + | |||
| + | ==See also== | ||
| + | |||
| + | {{AHSME box|year=1981|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:38, 28 June 2025
Problem 7
How many of the first one hundred positive integers are divisible by all of the numbers 
, 
, 
, and 
?
Solution
The least common multiple of , , , and is 
. There is only one multiple of 
between 
and 
, so the answer is 
-edited by coolmath34
See also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
