Find an integer between 100 and 200 such that each digit is odd and the sum of the cubes of the digit is equal to the original three-digit number. Hint: the "cube" of a number is that number multiplied by itself 3 times, so the cube of 2 is 2*2*2=8
Find an integer between 100 and 200 such that each digit is odd and the sum of the cubes of the digit is equal to the original three-digit number. Hint: the "cube" of a number is that number multiplied by itself 3 times, so the cube of 2 is 2*2*2=8
Hello, aznmartinjai!
With a little thinking, you can narrow down the choices.Find an 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.
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!