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. what number am i?

"i am a 3 digit number"

3 digit number = XYZ
x=hundreds
Y=tens
Z=ones

"divisible by 3"
(X+Y+Z)/3 = whole number

"my tens digit is 3 times as great as my hundreds digit"
Y=3X
therefore,(X+3X+Z)/3 = whole number

"the sum of my digits is 15"
X+Y+Z=15 or X+3X+Z=15 (from substitution of Y=3X) or Z=15-4X

"if you reverse my digits i am divisible by 6"
ZYX
which means it's divisible by 2!
X = 0, 2, 4, 6, or 8

plug X into here:
(X+3X+Z)/3 = whole number
Z=15-4X

see which ones work!

You would do better to call the School Subject MATH.

Sra

To solve this problem, we need to find a 3-digit number that meets several conditions:

Condition 1: The number is divisible by 3.
Condition 2: The tens digit is 3 times greater than the hundreds digit.
Condition 3: The sum of the digits is 15.
Condition 4: The number, when its digits are reversed, is divisible by both 6 and 3.

Let's start by considering Condition 1: The number is divisible by 3. For a number to be divisible by 3, the sum of its digits must be divisible by 3. Since the sum of the digits is given as 15 (Condition 3), we know it satisfies this condition.

Next, let's focus on Condition 2: The tens digit is 3 times greater than the hundreds digit. Let's represent the hundreds digit as 'x.' According to this condition, the tens digit would then be '3x'.

Now, let's consider Condition 3: The sum of the digits is 15. We know the units digit is divisible by 3 because the number itself is divisible by 3. The sum of the hundreds digit and the tens digit must equal 15 - the units digit.

So, x + 3x + (x + 3x) = 15 - (x + 3x)

Simplifying this equation, we have:
8x = 15 - 4x
12x = 15

Solving for x, we find x = 15/12 = 1.25. However, since we are looking for a whole number, we need to round down the result to the nearest whole number, which is 1.

Now that we have the value of the hundreds digit (x = 1), we can determine the tens digit (3x = 3) and the units digit (15 - x - 3x = 15 - 1 - 3 = 11).

Therefore, the 3-digit number meeting all the given conditions is 131, as the tens digit is 3 times greater than the hundreds digit (3 * 1 = 3), the sum of the digits is 15 (1 + 3 + 11 = 15), and when the digits are reversed, the number is divisible by both 6 and 3 (311 divided by 3 and 6 both have no remainder).