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Probability Level 3

Find the sum of all 3-digit natural numbers which contains at least one odd digit and at least one even digit.

1 solution

Kaustubh Miglani
Sep 21, 2015

To start off, we find the sum of all 3-digit natural numbers. So, because there are 900 natural 3-digit numbers starting from 100 and ending at 999, the sum is (100+999) \times (900)/2 = 494550. 2 Next, we find the sum of all 3-digit natural numbers that have only odd digits. There are five possible numbers that are odd for digits of the number, so there are 5 \times 5 \times 5 = 125 3-digit numbers that only have odd digits, starting from 111 and ending at 999. The sum of those numbers is (111+999) \times (125)/2 = 69375. 3 Then, we find the sum of all 3-digit numbers that only have even digits. There are five possible even numbers. However, since the number has 3 digits, 0 cannot be the hundreds digit, so there are (5-1) \times 5 \times 5 = 100 3-digit numbers that have only even digits, starting from 200 and ending at 888. The sum of those numbers is (200+888) \times (100)/2 = 54400. 4 Finally, we take the total sum of all 3-digit numbers and subtract by the sums of 3-digits numbers that only have odd or even digits to get our desired sum: the sum of 3-digit numbers that have at least one odd digit and one even digit. So, 494550 - 69375 - 54400 = 370775.