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Difference between revisions of "2012 AMC 12B Problems/Problem 5"

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==Solution==
 
==Solution==
  
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So, x+y=26, x could equal 15, and y could equal 11, so no even integers required here. 41-26=15. a+b=15, a could equal 9 and b could equal 6, so one even integer is required here. 57-41=16. m+n=16, m could equal 9 and n could equal 7, so no even integers required here, meaning only 1 even integer is required; A.
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Since, <math>x + y = 26</math>, <math>x</math> can equal <math>15</math>, and <math>y</math> can equal <math>11</math>, so no even integers are required to make 26. To get to <math>41</math>, we have to add <math>41 - 26 = 15</math>. If <math>a+b=15</math>, at least one of <math>a</math> and <math>b</math> must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from <math>26</math> to <math>41</math>. Finally, we have the last transition is <math>57-41=16</math>. If <math>m+n=16</math>, <math>m</math> and <math>n</math> can both be odd because two odd numbers sum to an even number, meaning only <math>1</math> even integer is required. The answer is <math>\boxed{(\text{bf}{A})}</math>. ~Extremelysupercooldude (Latex, grammar, and solution edits)
  
 
==Solution 2==
 
==Solution 2==

Revision as of 06:27, 29 June 2023

Problem

Two integers have a sum of $26$. when two more integers are added to the first two, the sum is $41$. Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$. What is the minimum number of even integers among the $6$ integers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

Since, $x + y = 26$, $x$ can equal $15$, and $y$ can equal $11$, so no even integers are required to make 26. To get to $41$, we have to add $41 - 26 = 15$. If $a+b=15$, at least one of $a$ and $b$ must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from $26$ to $41$. Finally, we have the last transition is $57-41=16$. If $m+n=16$, $m$ and $n$ can both be odd because two odd numbers sum to an even number, meaning only $1$ even integer is required. The answer is $\boxed{(\text{bf}{A})}$. ~Extremelysupercooldude (Latex, grammar, and solution edits)

Solution 2

Just worded and formatted a little differently than above.

The first two integers sum up to $26$. Since $26$ is even, in order to minimize the number of even integers, we make both of the first two odd.

The second two integers sum up to $41-26=15$. Since $15$ is odd, we must have at least one even integer in these next two.

Finally, $57-41=16$, and once again, $16$ is an even number so both of these integers can be odd.

Therefore, we have a total of one even integer and our answer is $\boxed{(\text{A})}$.

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 12 Problems and Solutions

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