2012 AMC 10B Problems/Problem 7: Difference between revisions
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<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 </math> | <math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 </math> | ||
==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. | ||
| Line 20: | Line 19: | ||
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 holes through common multiples of <math>3</math> and <math>4</math>, so <math>lcm[3,4]=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> | |||
==Solution 3== | |||
We can also use two variables to solve this problem. | |||
If the chipmunk stashed his acorns in <math>x</math> holes, and the squirrels stashed his acorns in <math>y</math> holes. | |||
We can get two equations: | |||
<math>x</math> = <math>y</math> + 4 and 3<math>x</math> = 4<math>y</math> | |||
When we solve this equation we can get <math>x</math> = 48, <math>y</math> = 44. | |||
So the answer is <math>\textbf{48(D)}</math> ~Mathlover1205 | |||
==Video Solution== | |||
https://youtu.be/hRlDVKgAv9U | |||
~savannahsolver | |||
==See Also== | |||
{{AMC10 box|year=2012|ab=B|num-b=6|num-a=8}} | |||
{{MAA Notice}} | |||
[[Category: Introductory Number Theory Problems]] | |||
Latest revision as of 17:48, 19 October 2025
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 holes through common multiples of 
and 
, so ![$lcm[3,4]=12$](http://latex.artofproblemsolving.com/1/a/d/1ad88f751508c871c757a7654863e5c575b85f58.png)
, 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 
Solution 3
We can also use two variables to solve this problem.
If the chipmunk stashed his acorns in holes, and the squirrels stashed his acorns in 
holes.
We can get two equations:
= + 4 and 3 = 4
When we solve this equation we can get = 48, = 44.
So the answer is ~Mathlover1205
Video Solution
~savannahsolver
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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