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1. The faces of a standard die are numbered 1, 2, 3, 4, 5, and 6 such that the sum of the numbers on any two opposite faces is 7. Tia writes one number on each vertex of the die such that the number on each face of the die is the greatest common divisor of the numbers at the four vertices of that face. What is the smallest possible sum of the eight numbers Tia writes?

 

 

2. Richard cuts a 4.5ftx7.5ftx11.25ft rectangular prism into congruent cubes without any part of the prism leftover. If he makes the cubes as large as possible, how many will he produce?

 May 11, 2019
 #1
avatar+8579 
+3

This is only for the second question:

 

We can multiply each dimension by  4  to make each dimension an integer. This will only scale the prism and the cubes up by  4....it won't change the number of cubes that can be fit into the prism.

 

So the new dimensions of the prism are  18 ft.  ×  30 ft.  ×  45 ft.

 

 

We can't cut the prism into cubes with sides length 18 ft, because 18 does not fit evenly into  30.

We can't cut the prism into cubes with sides length  9  ft, because  9  does not fit evenly into  30.

We can't cut the prism into cubes with sides length  2  ft, because  2  does not fit evenly into  45.

 

The side length of the cubes must fit evenly into 18, 30, and 45.

That is, the side length of the cubes must be a factor of 18, 30, and 45.

 

The maximum side length of the cubes is the greatest common factor of 18, 30, and 45.

 

The maximum side length of the cubes is  3 (ft.)

 

After splitting the prism into cubes with sides length 3 ft,

the prism will be 6 cubes high, 10 cubes wide, and 15 cubes deep.

 

the total number of cubes  =  6 * 10 * 15  =  900

 May 12, 2019
 #2
avatar+102417 
+2

Nice, hectictar   !!!!

 

 

cool cool cool

CPhill  May 12, 2019
 #3
avatar+200 
+1

Thanks Hectictar!

Given that you are so good at this, do you know how to solve the first problem? i've been having a lot of trouble on that one.

sudsw12  May 13, 2019

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