Difference between revisions of "2012 AMC 10B Problems/Problem 7"
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− | == Solution == | + | ==Solution 1== |
Let <math>x</math> be the number of acorns that both animals had. | Let <math>x</math> be the number of acorns that both animals had. | ||
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This is answer choice <math>\textbf{(D)}</math> | This is answer choice <math>\textbf{(D)}</math> | ||
+ | ==Solution 2== | ||
+ | Instead of an Algebraic Solution, we can just find a residue in the common multiples of <math>3</math> and <math>4</math>, so <math>lcm<cmath>3,4</cmath>=12</math>, the next largest is <math>12\cdot2=24</math>, the next is <math>36</math>, and so on, with all of them being multiples of <math>12</math>, now we can see that per every common multiple, we can see a pattern such as | ||
+ | <math>12=4\cdot3=3\cdot4</math> so <math>4-3=1</math> hole less. | ||
+ | <math>24=4\cdot6=3\cdot8</math> so <math>8-6=2</math> holes less. | ||
+ | <math>36=4\cdot9=3\cdot12</math> so <math>12-9=3</math> holes less. | ||
+ | <math>48=4\cdot12=3\cdot16</math> so <math>16-12=4</math> holes less. | ||
+ | |||
+ | So we see that <math>48</math> is the number we need which is <math>\textbf{48(D)}</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2012|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2012|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:43, 8 January 2018
Contents
Problem 7
For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?
Solution 1
Let be the number of acorns that both animals had.
So by the info in the problem:
Subtracting from both sides leaves
This is answer choice
Solution 2
Instead of an Algebraic Solution, we can just find a residue in the common multiples of and , so , the next largest is , the next is , and so on, with all of them being multiples of , now we can see that per every common multiple, we can see a pattern such as so hole less. so holes less. so holes less. so holes less.
So we see that is the number we need which is
See Also
2012 AMC 10B (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 | ||
All AMC 10 Problems and Solutions |
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