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Bunuel
Let A, M, and C be nonnegative integers such that A + M + C = 12. What is the maximum value of A*M*C + A*M + M*C + A*C ?


(A) 62
(B) 72
(C) 92
(D) 102
(E) 112

To maximise the equation, you need to maximise A*M*C part as it represents the highest multiplication.

Combinations of 12:

10 + 1 + 1 = 12 [10*1*1 = 10]
8 + 2 + 2 = 12 [ 8*2*2 = 32]
6 + 3 + 3 = 12 [ 6*3*3 = 54]
4 + 4 + 4 = 12 [4^3 = 64]

Therefore, 64 + 16 + 16 + 16 = 64 + 16(3) = 112
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Deconstructing the Question

We want to maximize

\(AMC+AM+MC+AC\)

subject to

\(A+M+C=12\)

with A, M, and C nonnegative integers.

A useful rewrite is

\((A+1)(M+1)(C+1)=AMC+AM+MC+AC+A+M+C+1\)

Since

\(A+M+C=12\)

the expression becomes

\(AMC+AM+MC+AC=(A+1)(M+1)(C+1)-13\)

So we only need to maximize

\((A+1)(M+1)(C+1)\)

Step-by-step

Let

\(x=A+1,\ y=M+1,\ z=C+1\)

Then

\(x+y+z=15\)

and we want to maximize

\(xyz\)

For three integers with fixed sum, the product is greatest when the numbers are as equal as possible.

So the best choice is

\(x=5,\ y=5,\ z=5\)

Then

\(xyz=5\cdot5\cdot5=125\)

Now go back to the original expression:

\(AMC+AM+MC+AC=xyz-13\)

So

\(125-13=112\)

This happens when

\(A=4,\ M=4,\ C=4\)

Checking:

\(4\cdot4\cdot4+4\cdot4+4\cdot4+4\cdot4=64+16+16+16=112\)

Answer: E
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