If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9

I got this question on MGMAT CAT i took yesterday.

According to the answer explanations, the answer should be D --> each statement ALONE is sufficient. My question is with statement (2). Below is my work for the problem:

X + 7 = n*9, where n is any integer

if n=2, X=11 and X/9 yields remainder of 2
if n=3, X=20 and X/9 yields remainder of 2
if n=-2, X=-16 and even though it is 2 units away from a multiple of 9, when -16/9, the remainder isn't 2 is it? Shouldn't the remainder be -7?

The problem does not explicitly state that X has to be positive. Am I interpreting this incorrectly? Needless to say, I picked answer A for this question.

andrew

typo...

if n=-1, X=-16 and even though it is 2 units away from a multiple of 9, when -16/9, the remainder isn't 2 is it? Shouldn't the remainder be -7?

sorry about the confusion...

(2) X + 7 is divisible by 9 (we can rewrite as (x+7)/9 = integer). so in this case x could be= 2, 11, 20, 29, 38....
when x=2, (2+7)/9=1, (11+7)/9=2, the question stem ask "what is the remainder when X is divided by 9?" so, when x=2, 2/9=0...remainder is 2, when x=11, 11/9=1...remainder is also 2. if you try x=20, 29, 38.. remainders will also be 2.. so, it is sufficient to answer the question.

by the way x (n) can not be 3 or -2 because when x=3, (3+7)/9= it's not integer, it can not be divi. by 9. The same as when x =-2, (-2+7)/9 = can not be divi. by 9

x cannot be 3 or -2, BUT n can be 3 or -2. can you double check?

Maybe i should rephrase or simplify my question.

Can X be negative?

In your analysis, you did not consider X to be negative. Please do that and see if statement (2) is sufficient.

Thanks a bunch for looking into this

andrew

x can not be negative.
(2) "X + 7 is divisible by 9" NEED TO BE INTEGER!! if x is negative then (x+7)/9 will NOT be integer and it will not satisfy the statement.
I did not use "Quotient and Remainder" formula: N=n*x+y. where N is dividend, n is divisor, x is quotient, and y is remainder. I just simply use x (in you case n is integer) as any integers. (as long as (an integer (x)+ 7 can be div.by 9.)

"X + 7 = n*9, where n is any integer". I am not sure in your original equation (formula) x+7=n*9 is correct , since (x+7) is dividend, n is divisor, 9 is quotient in this case, and no remainder.
so if you use (X+7)=n*9 as the guideline, then 9 will the quotient. in statement 2, 9 should be divisor. so the equation should look like (X+7)=9*n+0. n is quotient and can be any integer. for ex: x=2, then (2+7)=9*1+0, n=1. if x=-2, then (-2+7)=9*n, n=-5/9 is not integer. Remember in statement 2 -----(x+7)/9 need to integer.

i do not agree with your statement: "if x is negative then (x+7)/9 will NOT be integer and it will not satisfy the statement."

For example, if x=-16, then
(x+7)/9 = (-16+7)/9 = -9/9 = -1
-1 is an integer.

Likewise, if x=-25, then
(x+7)/9 = (-25+7)/9 = -18/9 = -2
-2 is an integer.

Also, if x=-34
(x+7)/9=(-34+7)/9=-27/9=-3
-3 is an integer.

I have listed 3 examples above where x is a double-digit number that satisfies statement (2), x + 7 is divisble by 9. However, when these x values are divided by 9, the remainder is not 2. The remainder is -7. Since the remainder can either be 2 or -7, this statement is insufficient.

My question of whether x can be negative does not mean that x can be any negative integer. Your examples of choosing x=-1 and x=-2 are examples that do not satisfy statement (2), but that does not mean no negative x value can satisfy statement (2). Please be clear on that. Also, according to the question stem, x needs to be a double-digit number.

You reason that x cannot be negative because if x is negative then (x+7)/9 will not be an integer. I have shown 3 examples above that proves the contrary. Please provide another reason.

Any help from other individuals would be greatly appreciated. Thanks

The question setter has ,no doubt, assumed the two digit integer to be positive.Getting into the mind of the question setter is key to the 'correct' answer.
Some question setters lack mathematical genius themselves but do have the remarkable prowness of stimulating it.

There seem to be two issues being discussed in this thread. One issue is whether (x+7)/9 can yield an integer if x is negative. As one poster pointed out, yes it can.

The second question is whether the premises of the problem allow x to be negative. They do not. We are told that a and b are the digits of the two-digit number X. Digits by definition, are the integers 0 - 9. Since a and b are both positive (or 0), X is also positive or 0.

More generally, the GMAT is not likely to ask about remainders when the quotient is negative.

The second question is whether the premises of the problem allow x to be negative. They do not. We are told that a and b are the digits of the two-digit number X. Digits by definition, are the integers 0 - 9. Since a and b are both positive (or 0), X is also positive or 0. " How are the digits of negative numbers defined? To presume the digits a and b can't be both negative is erroneous. Any two digit number can be represented algebricly as 10a +b or 10b+a.a &b will both be negative digits if x is negative.

'More generally, the GMAT is not likely to ask about remainders when the quotient is negative" is an assumtion that may be true more often than not but can be perilous depending on the mind of the question-setter.

Shaj,

It's good to be skeptical! But I have been teaching GMAT for ~14 years and feel pretty confident on this one. That said, if this problem came up on the real exam they probably would have stated that X was positive to avoid any potential ambiguity. Let me demonstrate why negative digits are a problem:

What are the digits of the number 27?

I would answer 2 and 7. 2 is the tens digit and 7 is the ones digit. And 2 * 10 + 7*1 = 27, so it checks.

But according to your system, we would be equally justified in saying the digits of 27 are 3 and -3. 3 is the tens digit and -3 is the ones digit. This works out OK because 3 * 10 + -3*1 = 27! There is nothing really wrong with this, except that it is confusing. So to avoid ambiguity, we usually consider the digits of any number to be positive or 0. This does raise the question, as you say, what are the digits of a negative number? Well to keep things relatively simple, we keep ALL the digits positive (or 0). And then we stick a negative sign on the front (not assigned to any particular digit). So we can assume in this problem that a and b are positive. For the GMAT to then say that x is a digit number with (presumed) positive digits a and b, and for them to mean that x could be negative, would be terribly misleading. The GMAT does like to make things hard, but NOT through the use of ambiguous definitions. A very similar argument can be made about remainders. It is safe to assume here that a, b and x are positive. On the actual GMAT that likely would have been made explicit.

The important thing anyway is not getting this question right, but learning something so you can get the next question right. All the key points to learn here are what follows from the solution once you assume that x and a and b are all positive. The rest is mostly a distraction (though interesting!).

Jonathan;
It good to be confident, indeed!!!. Yes! >14 years of teaching GMAT is no joke, all due respects. Mathematics, the queen of all sciences , is no respector of anyone/anything, save logic/rationale or perhaps 'distractions'.

I feel it worthwhile to clarify the rationale of depicting two digits numbers algebricly.The logic is 'a' & 'b' are digits that both take either +ve or -negative values for postive and negative numbers respectively. 'a' & 'b' can't take opposing signs at any instant as they are the digits of an integer.

I do wish all users of this forum benefit the most from such discussions and be wary that the GMAT is evolving with time and adapting itself to make the best out of these 'distractions' in an effort to get the best propestive managers into mangement schools.

And finally, this question had contributed well to stimulate mathematical genius.

yes, good to be on the look out for new things. :)