I am a 3 digit number dividible by 3. My tens digit is 3 times as great as my hundreds digit,
and the sum of my digits is 15.
If you reverse my digits, I am divisible by 6, as well as by 3.
What number am I?
267
To find the number, let's break down the given information step by step.
We are given that the number is divisible by 3 and has 3 digits. So, let's consider all the possible 3-digit numbers that are divisible by 3: 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, and so on.
Next, we're given that the tens digit is 3 times as great as the hundreds digit. This means that the hundreds digit can only be 1, 2, or 3, as 4 or any higher number would result in a tens digit greater than 9. So, let's consider the possible numbers based on this criterion: 102, 105, 112, 114, 123, 129, and so on.
Now, we know that the sum of the digits is 15. Let's calculate the sum of the digits for each of the remaining numbers:
- 102: 1 + 0 + 2 = 3
- 105: 1 + 0 + 5 = 6
- 112: 1 + 1 + 2 = 4
- 114: 1 + 1 + 4 = 6
- 123: 1 + 2 + 3 = 6
- 129: 1 + 2 + 9 = 12
From the remaining numbers, only 105, 114, and 123 have a sum of digits equal to 15.
Now, let's check the divisibility criteria. Reversing the digits of each number:
- 105 becomes 501 (divisible by 3 and 6)
- 114 becomes 411 (divisible by 3 but not by 6)
- 123 becomes 321 (divisible by 3 and 6)
Therefore, the number that satisfies all the given criteria is 123.