+0  
 
+1
752
6
avatar+936 

Answer: 70

 

I tried this question by drawing pictures, but it didn't work out.

 Jul 27, 2019
 #1
avatar+6244 
+3

\(\text{It can't be 70}\\ A_{board} =32^2 = 1024 ~sq~in\\ A_{photo}=3\cdot 5 = 15 ~sq ~in\\ \dfrac{1024}{15} = 68+\dfrac{4}{15}\\ \text{so with no overlapping 68 is the maximum possible, and is probably not attainable}\)

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 Jul 27, 2019
 #2
avatar+9460 
+4

As Rom said, there is definitely no way to put 70 pictures on the board.

 

68 pictures is the maximum possible, and here is a way to do that:

 

 

However, if you can't rotate the pictures 90 degrees, then the maximum possible is

 

floor( 32 / 3 )  *  floor( 32 / 5 )   =   10 * 6   =   60

 Jul 27, 2019
 #3
avatar+128079 
+2

Nice, hectictar   !!!!

 

 

cool cool cool

CPhill  Jul 27, 2019
 #4
avatar+936 
+3

I don't know guys. Out of all the questions I've looked at made by this organization, their answers are always correct. Also, this was on a math contest taken by all middle schoolers in the U.S., so if the answer is wrong, then a student has to have already told them and they would have fixed it already. I also got 68, but these questions tend to have a creative way to solving them. Still, you guys have really great solutions and I agree with your reasoning why 68 is the maximum possible.

dgfgrafgdfge111  Jul 28, 2019
edited by dgfgrafgdfge111  Jul 28, 2019
edited by dgfgrafgdfge111  Jul 28, 2019
 #5
avatar+9460 
+4

Hmmm....even if you cut the pictures into 1 inch by 1 inch squares, there's just not enough square inches on the board to fit 70 pictures worth.

 

number of square inches on the board  =  32 * 32  =  1024

 

number of square inches of 70 pictures  =  70 * 3 * 5  =  1050

 

Are you sure it says  70  and not  60  ?

 

If so, maybe you can be the one to report the error in this question. Yes it is weird and unlikely they'd have it wrong, but it's not impossible.

hectictar  Jul 28, 2019
 #6
avatar+6244 
+3

As an aside if I were writing this problem I would not use photos.

No one is going to rotate photos 90 degrees to display them.

Using tiles would have made the problem far more sensible with out affecting any of the underlying math.

Rom  Jul 28, 2019

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