Is a four digit number "abcd" divisible by 9?

Hi All,

This is a layered question, but not too terrible. If you recognize that the two Facts really come down to the values of A and B, then you can answer this question without too much time.

We're told that the ABCD represent digits in a 4-digit number. We're asked if ABCD is divisible by 9. This is a YES/NO question.

The question is built around the concept of "the rule of 9": if the DIGITS of the number SUM to a value that is divisible by 9, then the larger number is also divisible by 9. If the digits do NOT SUM to a value that is divisible by 9, then the larger number also IS NOT divisible by 9.

For example:

18 is divisible by 9 because 1+8=9, which is divisible by 9.
17 is NOT divisible by 9 because 1+7 = 8, which is NOT divisible by 9

Fact 1: (A^B)(B^C) is divisible by 9

If we can make either A^B or B^C divisible by 9, then it doesn't matter what the other parentheses equals. As it stands, this Fact tells us NOTHING about the value of D. That value will impact whether ABCD is divisible by 9 or not.
Fact 1 is INSUFFICIENT

Fact 2: (A+B)(C+D) is divisible by 9

Again, if we can make either (A+B) or (C+D) divisible by 9, then it won't matter what the other parentheses equals.

IF....
A=B=C=D = 9
(18)(18) is divisible by 9
9+9+9+9 = 36, which is divisible by 9
The answer to the question is YES

IF...
A=3
B=6
C=D=1
(9)(2) is divisible by 9
3+6+1+1 = 11, which is NOT divisible by 9
The answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, notice how BOTH Facts give us a parentheses with A and B in it....
(9^9) and (9+9) are divisible by 9
(3^6) and (3+6) are divisible by 9
BOTH of the TESTs that I used in Fact 2 will ALSO "fit" Fact 1, giving us a YES and a NO answer.
Combined, INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hello vipulgoel.
Yeah, you are absolutely right. This is one of those stupid mistakes when I forget about what question asks (
I do a lot of such things during CATs, but make such one on the forum, this is incredible carelessness. Looks like this is my new record

ya , It happens with lot of us

I am wondering if we can apply divisibility by 9 rule here. If yes, then for abcd to be divisible by 9, a+b+c+d must be divisible by 9.

I tried,but none of the expression gives me that and ended up with E. I don't prefer value approach here.
Can anyone give more concrete explanation.


Thanks,
Gaurav

Hello GauravSolanky
(1) (a^b)(b^c) is divisible by 9
we know nothing about d
Insufficient

(2) (a+b)(c+d) is divisible by 9
if we want this expression to be divisible by 9 one of the part (a+b or c+d) should be divisible on 9
the easiest case: all numbers are equal to 9 -- abcd is divisible by 9
a+b = 3+2 = 5 and c+d = 4+5 = 9 -- abcd is not divisible by 9
Insufficient

1+2)
If all numbers are divisible by 9 then both statements are true and abcd is divisible by 9
and if a=3, b=2, c=4, d=5 then both statements are true and abcd is not divisible by 9

The hardest part is to pick numbers for both statements
we can easily find the case with all 9
and for negative case we should try to pick numbers that will give satisfy such conditions:
9 as a factor of this part (a^b) --> for first statement be true
9 as a factor of (c+d) --> for second statement be true
and (a+b) shouldn't be divisible by 9 --> for negative result of the question

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