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What is the integer between 100 and 200 such that each digit is odd and the sum of the cubes of the digits is equal to the original three-digit number?
Wiki User
∙ 10y ago
With a little thinking, you can narrow down the choices.
Since the number is between 100 and 200, the first digit is "1".
The other digits must be odd: 1, 3, 5, 7, 9
But 7³ = 343 and 9³ = 729 . . . too large.
So the number is comprised of 1s, 3s, and/or 5s.
If the number has no 5s, the largest is 133 . → . 1³ + 3³ + 3² .= .55
. . Hence, the number contains one 5.
There are only four choices: 115, 135, 151, 153.
. . 115 . → . 1³ + 1³ + 5³ .= .127
. . 135 . → . 1³ + 3³ + 5³ .= .153
. . 151 . → . 1³ + 5³ + 1³ .= .127
. . 153 . → . 1³ + 5³ + 3³ .= .153 . ← .There!
Wiki User
∙ 10y ago
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