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1997 PMWC Problems/Problem I14

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Problem

If we make five two-digit numbers using the digits 0, 1, 2, \ldots 9 exactly once, and the product of the five numbers is maximized, find the greatest number among them.

Solution

This is a greedy algorithm question. Let a_{1}, b_{1} be the digits of the first number, etc. Without loss of generality let 10a_1 + b_1 be the greatest number. Then we want to maximize the quantity

(10a_1 + b_1)(10a_2 + b_2) \cdots (10a_5 + b_5) =10^5 a_1a_2a_3a_4a_5 + 10^4(b_1a_2a_3a_4a_5 + \ldots) + \ldots + b_1b_2b_3b_4b_5

The greedy algorithm quickly tells us that the first digits of the numbers should be 9,8,7,6,5, so a_1 = 9. Now, look at the coefficient of 10^4. The product a_2a_3a_4a_5 is less than any of the other terms (which all contain the maximal a_1 = 9), so by the greedy algorithm, we should make b_1 as small as possible. Hence b_1 = 0, and our answer is 90.

See also

1997 PMWC (Problems)
Preceded by
Problem I13
Followed by
Problem I15
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
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