Is a 3-digit integer divisible by 3?

(1) The hundreds' digit of the number is equal to the tens' digit of the number.

(2) The units' digit of the number is 6 greater than the tens' digit of the number

OA

A little confused

I'll start by naming the integers A B C

(1) if 100's digit is = 10's digit then A = B so it now becomes A A C unit place is not given so INSUFF

(2) if unit digit is 6 greater than the 10's digit the number will read as A B 6+B since 100's digit is not given it still is INSUFF (to me)

BOTH together then it will read something like this A A 6+A

so the numbers will be either

116 ,

227 or 339 to me it is still INSUFF and i picked E

where am i going wrong ??

Sorry about posting in such a short time My exam is very very close!!!!

First # should be 117, which is divisible by 3 and second number should be 228, which is also divisible by 3. So all 3 numbers possible: 117, 228 and 339 are divisible by 3.

A number is divisible by 3 if the sum of its digits is divisible by 3. So 3-digit integer \(abc\) will be divisible by 3 if \(a+b+c=3k\), where k is integer or to make it simpler if \(a+b+c\) is a multiple of 3.

(1) The hundreds' digit of the number is equal to the tens' digit of the number --> \(a=b\) --> \(a+b+c=b+b+c=2b+c\). Now, \(2b+c\) may or may not be a multiple of 3, hence this statement is insufficient.

(2) The units' digit of the number is 6 greater than the tens' digit of the number --> \(c=b+6\) --> \(a+b+c=a+b+b+6=a+2b+6\). Now, \(a+2b+6\) may or may not be a multiple of 3, hence this statement is insufficient.

(1)+(2) As from (1) \(a=b\) and from (2) \(c=b+6\) then \(a+b+c=b+b+b+6=3b+6=3(b+2)\). As \(3(b+2)\) has 3 as multiple hence it's divisible by 3. Sufficient.

Answer: C.