If x is a positive three-digit integar what is the tens digit of x

I am a bit confused here.

Can we have 0 in the tens place. Because 9 times of 0 is also 0.

So, we can have two numbers 891 and 900 both satisfying the above two statements.

If my understanding is correct, answer should be E. else I am good with C.

I will wait for experts to throw some light on it.

I went with C for donerat of the same reasoning as above. Howevet, I also am curious whether 090 would be considered a three digit integer.

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No. 090 is not a three digit number. But 900 is. Lets see what suggestions come.

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A=108,117,126......
B=901,911,921........981,991
C=981


In 900, hundreds digit is 9 which is not 9 times of the unit digit (zero).

(0x9=0 not 9)



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But 0*9 = 0
So 900 is not possible. (IF the answer choice was hundreds digit and unit digit are divisible by 9 or multiples of 9 then 900 would work. But we are told that (Number=xyz) x=9z. Either both are zero(no longer a three digit number) or x=9 and z=1. 9 and 0 is not an option.

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This is quite a simple question, but the trick is to not fall victim to statement carryover.

When evaluating statement 2 FORGET statement 1.

I hope it helps someone.


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Statement two says that the hundreds digit of x is 9 times the units digit. 999 does not fulfill that requirement. 981 is the number.

The question is not a +700 GMAT question.

The solution is pretty straightforward:
1) 9*12 = 108 & 9*111 = 999 --> 111-12 + 1 = 100 # of possible 3 digit # multiple of 6
2) say that X is 9y1, so 10 possible solution

1&2
Apply divisibility rule of 9: the sum of the digit must be divisible by 9 , so 9+y+1 = 18 --> y=8 --> x=981 exist only one solution

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

\(x = 100a + 10b + c\) where \(a\), \(b\), \(c\) are integers such that \(1 \le a \le 9\), \(0\le b,c \le 9\).
The question asks what the value of \(b\).

There are 4 variables and 1 equation. Thus E is the answer most likely.

Condition 1) : \(a + b + c\) is a multiple of 9.
This is not sufficient, since there are many combination of \(a\), \(b\) and \(c\).
For example, \(a = b = c = 3\) and \(a = b = c = 9\).

Condition 2) \(a = 9c\)
We have \(a = 9\) and \(c= 1\) from this condition.
But we can not get anything else about \(b\).

Condition 1) & 2)
There is a unique solution for these conditions, which is \(a = 9\) and \(c= 1\) and \(b = 8\), since \(a + b + c\) is a multiple of 9.
This is sufficient.

Therefore, the answer is C).

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.

*******************************
let three digit number X=ABC (ABC are digits), B=?

Statement 1
X is multiple of 9, by divisibility rule A+B+C should be divisible by 9 but we do not have any idea about the values of ABC. Insufficient.

Statement 2
A=9C, C can take value from 0 to 9. let us check
C=0 A =0 number will be two-digit so out
C=1 A= 9 looks OK
C=2 A=18 it will become four-digit, so out

We know X=9B1 but still no clue about B. Insufficient.

Statement 1 and 2 together

A+B+C should be divisible by 9 (statement 1)
X=9C1 (statement 2)

9+B+1 =9I, only B=8 satisfies

The answer is C.

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