During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

Difference between revisions of "2012 AMC 10B Problems/Problem 9"

(Created page with "== Problem 9 == Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of th...")
 
(Solutions)
Line 10: Line 10:
 
== Solutions ==
 
== Solutions ==
  
Lets say that all 6 integers added are : a,b,c,d,e, and f.
+
Lets say that all 6 integers added are : <math>a,b,c,d,e,</math> and<math> f</math>.
  
If a+b=26
+
If <math>a+b=26</math>
  
and a+b+c+d=41
+
and <math>a+b+c+d=41</math>
  
 
Then,
 
Then,
c+d=15
+
<math>c+d=15</math>
  
 
Also,  
 
Also,  
  
a+b+c+d+e+f=57
+
<math>a+b+c+d+e+f=57</math>
  
a+b+c+d=41
+
<math>a+b+c+d=41</math>
  
 
Then,
 
Then,
e+f=16
+
<math>e+f=16</math>
  
  
 
So
 
So
  
a+b=26
+
<math>a+b=26</math>
  
c+d=15
+
<math>c+d=15</math>
  
e+f=16
+
<math>e+f=16</math>
  
a,b,e,f can be all odd since odd + odd= even. And the sum of the two respective pairs are even.
+
<math>a,b,e,f </math>can be all odd since odd + odd= even. And the sum of the two respective pairs are even.
  
However, either c or d has to be even to get a odd sum.
+
However, either<math> c</math> or <math>d</math> has to be even to get a odd sum.
  
 
Therefore, there is <math>\boxed{1}</math> even integer
 
Therefore, there is <math>\boxed{1}</math> even integer

Revision as of 17:52, 24 February 2012

Problem 9

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

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\5$ (Error compiling LaTeX. ! Undefined control sequence.)

Solution


Solutions

Lets say that all 6 integers added are : $a,b,c,d,e,$ and$f$.

If $a+b=26$

and $a+b+c+d=41$

Then, $c+d=15$

Also,

$a+b+c+d+e+f=57$

$a+b+c+d=41$

Then, $e+f=16$


So

$a+b=26$

$c+d=15$

$e+f=16$

$a,b,e,f$can be all odd since odd + odd= even. And the sum of the two respective pairs are even.

However, either$c$ or $d$ has to be even to get a odd sum.

Therefore, there is $\boxed{1}$ even integer

OR

$\textbf{(A)}$

Invalid username
Login to AoPS