3 digit # divisible by 3. tens digit is 3 x as great as 100's digit , sum digit 15 Reverse digit x 6 & 3?

Question

Once again receive this worksheet with no knowledge of divisibility from new teacher only in 6th grade never review or given instruction on how to do this. Need detailed anwsers so that we can understand and be able to use extensive response.

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 by 3. What number am I

Answer 1

So we have a 3 digit number. Let the digits be:

a = 100's place
b = 10's place
c = units.

a+b+c = 15

b = 3a

a+b > 5

Now, a must be an integer. a can only be 1, 2 or 3. If A is one, then b is 3 and since the sum of all three digits is 15, it can't exist.

a = 2, b = 6, then b = 7, so that a + b +c = 15. This sums up those terms

a = 3, b = 9, c = 3, also fits conditions. I don't understand "Reverse digit x6 &3 "

Answer 2

OK, the first part makes sense, but the last part "sum digit 15..." doesn't make any sense. Is that the way it's written on the worksheet? If not, could you post the problem exactly as it's written on the worksheet?

Ah, now it makes sense, after reading the actual problem and John T's first answer, 267, would be correct.

Answer 3

There is no way to arrange these number to make a number that is perfectly divisible by 7. However, if you are allowed to use a number twice, 161 is divisible by 7. 7 x 23 = 161