AMC 10 Step by Step

2004 AMC 10B Problem 5

P1-10~2 minPrint

In the expression c⋅ab−dc\cdot a^b-d , the values of aa , bb , cc , and dd are 00 , 11 , 22 , and 33 , although not necessarily in that order. What is the maximum possible value of the result?

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Key insight

Put the 0 where it does no damage: as the number being subtracted. Then try the six ways to place 1, 2, 3; one times 3 squared gives 9.

Solution

The 00 does the most damage, so decide where it goes first.

  1. If a=0a = 0 or c=0c = 0, the product c⋅abc \cdot a^b is 00 and the result is 0−d0 - d, which is negative. Bad.
  2. If b=0b = 0, then a0=1a^0 = 1, and the result is c−dc - d. With cc and dd from 1,2,31, 2, 3, the best is 3−1=23 - 1 = 2. Small.
  3. If d=0d = 0, nothing is subtracted, and the result is just c⋅abc \cdot a^b. This is clearly the best place for the 00.

Now place 1,2,31, 2, 3 into a,b,ca, b, c. There are only six ways, so try them all:

ccaabbc⋅abc \cdot a^b
1133221⋅32=91 \cdot 3^2 = 9
1122331⋅23=81 \cdot 2^3 = 8
2233112⋅31=62 \cdot 3^1 = 6
2211332⋅13=22 \cdot 1^3 = 2
3322113⋅21=63 \cdot 2^1 = 6
3311223⋅12=33 \cdot 1^2 = 3

The biggest value is 99, with a=3a = 3, b=2b = 2, c=1c = 1, d=0d = 0.

The answer is (D) 9\boxed{\textbf{(D)}\ 9}.

Why this works

When there are only a handful of arrangements, a quick organized list beats clever guessing. A few facts help you see why the winner wins: a 00 belongs where it is subtracted, a 11 is wasted as a base since 11 to any power is 11, and 32=93^2 = 9 is a bit bigger than 23=82^3 = 8.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Decide where the 00 goes. If a=0a = 0 or c=0c = 0, the product vanishes and the result is −d-d, negative. If b=0b = 0, then ab=1a^b = 1 and the result is c−d≤3−1=2c - d \le 3 - 1 = 2. So the best use of 00 is d=0d = 0, which also removes the subtraction entirely.

Now distribute 1,2,31, 2, 3 among a,b,ca, b, c to maximize c⋅abc \cdot a^b. Test the six orders:

  • c=1c = 1: 32=93^2 = 9 or 23=82^3 = 8.
  • c=2c = 2: 2⋅31=62 \cdot 3^1 = 6 or 2⋅13=22 \cdot 1^3 = 2.
  • c=3c = 3: 3⋅21=63 \cdot 2^1 = 6 or 3⋅12=33 \cdot 1^2 = 3.

The largest is 1⋅32=91 \cdot 3^2 = 9, from (a,b,c,d)=(3,2,1,0)(a, b, c, d) = (3, 2, 1, 0).

The answer is (D) 9\boxed{\textbf{(D)}\ 9}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2004 12B #2.