2006 AIME II 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 $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting).

Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \cdot 2 = 162$ . Therefore, there are $999 - (99 + 162) = 738$ such ordered pairs.

Solution2

Assume a and b are both 3 digit numbers. Cal their digits c,d,e and f,g, and h. We have

cde

+fgh ____ 1000

$E$ and $h$ must add up to 10, $d$ and $g$ must add up to 9, and $c$ and $f$ must add up to 9. Since none of the digits can be 0, there are $9x8x8=576$ possibilites if both numbers are three digits.

There are two other scenarios. A and b can be a three digit number and a two digit number, or a three digit number and a one digit number. For the first scenario, there are $9x8x2=144$ possibilities (Since you can interchange whether $a$ or $b$ is three digits) and for the second case there are $9x2=18$ possibilities. Thus, thus total possibilities for $(a,b)$ is $576+144+18=738.$

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

Contents

Problem

Solution

Solution2

See also