Difference between revisions of "2012 AMC 8 Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | To find the answer to this problem, we need to find the least common multiple of <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math> and add <math>2</math> to the result. The least common multiple of the four | + | To find the answer to this problem, we need to find the least common multiple of <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math> and add <math>2</math> to the result. The least common multiple of the four numbers is <math>60</math>, and by adding <math>2</math>, we find that that such number is <math>62</math>. Now we need to find the only given range that contains <math>62</math>. The only such range is answer <math>{\textbf{(D)}</math>, and so our final answer is <math> \boxed{\textbf{(D)}\ 61\text{ and }65} </math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=14|num-a=16}} | {{AMC8 box|year=2012|num-b=14|num-a=16}} |
Revision as of 18:32, 22 December 2012
Problem
The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?
Solution
To find the answer to this problem, we need to find the least common multiple of , , , and add to the result. The least common multiple of the four numbers is , and by adding , we find that that such number is . Now we need to find the only given range that contains . The only such range is answer ${\textbf{(D)}$ (Error compiling LaTeX. ! Missing } inserted.), and so our final answer is .
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |