i am a 3-digit number divisible 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 3
To solve this problem, we need to use some basic principles of divisibility and number manipulation.
Let's start by considering the sum of the digits. We know that the sum of the digits is 15. We can represent the 3-digit number as ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit.
From the given information, we can write two equations:
1. A + B + C = 15 (sum of digits equation)
2. B = 3A (tens digit 3 times greater than hundreds digit)
To solve these equations, let's substitute B in the first equation with 3A (from equation 2):
A + 3A + C = 15
4A + C = 15
Now let's move on to the divisibility conditions. The number is divisible by 3, so the sum of its digits must also be divisible by 3. We already have the equation A + B + C = 15. Since we know A + B + C = 15, we also know that 4A + C = 15.
The second divisibility condition is that when the digits are reversed, the number should be divisible by both 3 and 6. Since we have a 3-digit number, reversing its digits gives us a new number CAB.
To check if CAB is divisible by 6, we need to make sure it is both divisible by 3 (from the given information) and even (divisible by 2). If C is even, then CAB will be divisible by 6.
Now, let's try to solve for A, B, and C.
From the equation 4A + C = 15, we can consider all possible values for A and C that satisfy the equation and other given conditions.
If we let A = 1, then C = 11, which is not a valid digit.
If we let A = 2, then C = 7, which is a valid digit.
If we let A = 3, then C = 3, which is not a valid digit.
If we let A = 4, then C = -1, which is not a valid digit.
Therefore, the only valid solution is A = 2, B = 6 (3A), and C = 7.
Hence, the 3-digit number divisible by 3, with the tens digit 3 times greater than the hundreds digit, and a sum of digits equal to 15 is 267.