2006 AIME I Problems/Problem 7

Problem

Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

Solution

There are $\frac{1000}{10} = 100$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities for $a$ are when $a$ or $b$ have a 0 in the tens digit, or if the tens digit of $a$ is 0 or 9. Excluding the hundreds, which were counted above already, there are $9 \cdot 2 = 18$ numbers in every hundred numbers that have a tens digit of 0 or 9, totaling $10 \cdot 18 = 180$ such numbers. However, the numbers from 1 to 9 and 991 to 999 do not have 0s, so we must subtract $18$ from that to get $162$ . Therefore, there are $1000 - (100 + 162) = 738$ such ordered pairs.

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2006 AIME I Problems/Problem 7

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