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# First question?

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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.

Jul 7, 2023

#1
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If neither a nor b has a zero digit, then the largest digit that either a or b can have is 9. Therefore, a must be between 1 and 999, inclusive, and b must be between 1 and 901, inclusive. Note that (2,998) and (998,2) should be counted as two distinct solutions.

We can count the number of solutions by considering the ones digit, the tens digit, and the hundreds digit of a and b separately. If the ones digit of a is 9, then the ones digit of b must be 1, so there is only 1 solution for this case. If the ones digit of a is 1 through 8, then the ones digit of b can be anything from 0 through 8, so there are 10 solutions for this case. Similarly, there are 10 solutions for each of the cases where the tens digit of a is 9, 1 through 8, and 0, and there are 10 solutions for each of the cases where the hundreds digit of a is 9, 1 through 8, and 0. Therefore, there are a total of 1+10⋅3⋅10=450​ solutions.

Jul 7, 2023
#2
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1 + 999 =1000
2 + 998 =1000
3 + 997 =1000
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500 + 500 =1000 [That is 500 pairs]

There are 180 integers between 1 and 999 inclusive, with at least one "0" in them:
Therefore: 180 / 2 =90 - pairs that should not be included

So: 500 pairs - 90 pairs =410 - ordered pairs.

Jul 7, 2023