2017 AMC 10B Problem 6

Question

What is the largest number of solid

2

-in

\times

2

-in

\times

1

-in blocks that can fit in a

3

-in

\times

2

-in

\times

3

-in box?

3

4

5

6

7

Start with the volume upper bound

Then check that four blocks really can be arranged

The volume of the large solid object is

3\cdot 3\cdot 2 = 18

and volume of the smaller object is

2\cdot 2\cdot 1=4.

This means we can fit at most

4

of the small objects.

We can make this happen by putting

3

of the small objects in a

3 \times 2 \times 2

rectangular prism, and then we have a

3 \times 2 \times 1

space left where we can place one small object.

Thus, the correct answer is B .

Problem 6 in other years