The difference 942 − 249 is a positive multiple of 7. If a, b, and c a

The difference between the two digits (abc − cba) = \(100a+10b+c - 100c-10b-a = 99a-99c = 99(a-c)\)
As 99 is not divisible by 7, then \((a-c)\) must be divisible by 7 (and b has no effect on the overall outcome)

\((a-c)\) are only 2 possible values: \((9-2)\) and \((8-1)\) ;as a,b,c are non zero; and a must be > c to keep the positive sign.
b has \(9\) possible values (all except 0)

So the total number of possible outcomes = \(2*9 = 18\)
E

Given: The difference 942 − 249 is a positive multiple of 7.

Asked: If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?

(100a + 10b + c) - (100c + 10b + a) = 99(a-c)
Since 99 is not a multiple of 7, (a-c) should be a multiple of 7.

There are 2 possibilities = {(8,1),(9,2)}
There are 9 possibilities for b

Total such numbers = 2*9 = 18

IMO E

The numbers are 100a + 10b +c and 100c + 10b + a.
After taking the difference, we are left with 99(a-c).
Now for this number to be a multiple of 7, a-c = 7 or 0.
for (a-c) = 7, there are two possibilities ---> (9,2) and (8,1)
and for a-c = 0, a = c, so there are 9 possibilities that are 1 to 9.

The value of b does not matter and since b cannot the value 0, so there 9 possibilities for b in each case.

There are a total of 11 cases so the total number of numbers possible are 9*11 = 99.

IMO, answer is C

given
100a+10b+c-100c-10b-a ; 99(a-c)
for 99(a-c) to be divisible (a-c) has to be divisible by 7 and a,b,c are all single digit non zero ;
only possibility of (a-c) = 7 ; ( 9-2) & (8-1) ; 2 choices of a & c ; for b which is middle term we have 9 options ; so total possible values; 9*2 ; 18
IMO E

One question on this - If a, b and c are given as different alphabets, should we consider them as distinct or not?

No, You can't consider them distinct without being given that they are different/distinct integers or something such as a≠b≠c or a>b>c

By this method, the number 252, 343, 515 etc. all qualify to represent 'abc'.
Let's take 252 as abc.
 cba = 252.
 abc - cba = 0.
But the question wants the difference to be a positive multiple of 7.
Now, 0 is neither positive nor negative.
Thus, when abc = 252, then abc – cba is not a positive multiple of 7.
 The condition a = c is not valid.
Only (a,c) = (9,2) and (8,1) is possible.
In both cases b can take 9 possible values as b ≠ 0.
 2 x 9 = 18 possible choices.

Can we solve the problem this way?

942 = 3^2 2^2 2^1

Total Factors = (2+1) * (2+1) * (1+1) = 18

942 is given as an example to show what the question means. The number of factors of 942 has nothing to do with the solution.

The difference of a 3-digit number and its reverse is always divisible by 99.

Now, 99 does not have a factor of 7 in it.

Expressing the 3-digit number as xyz and its reverse as zyx, we have their difference to be 99 (x-z) { because xyz can also be written as 100x + 10y + z and zyx as 100z + 10y + x}. So, the (x-z) has to have a factor of 7 if the difference has to be a multiple of 7 as the question says.

Considering that all the digits involved here are non-zero, (x-z) can be a factor of 7 in only two ways i.e. x=9 and z = 2 AND x=8 and z=1.
For each of these cases, y can be dealt in 9 ways i.e. y can be any digit from 1 to 9. Therefore, we have a total of 2*9 = 18 numbers.

The correct answer option is E.


Hope that helps!
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One query, what if the questions says that a,b,c are distinct integers?
VeritasKarishma mam can you please explain.

The only change in that case would be that b can take only 7 values, not 9.

(a, c) can be (9, 2) or (8, 1). For each option, b can take 7 distinct values. a and c are anyway distinct.
So answer will be 2*7 = 14

Why can we ignore the possibility of (a-c) = 0 in this case? As 0 should be divisible by 7, this would mean that a=c and that the resulting numbers abc and cba were identical. Therefore, the difference would always yield 0 as well.

Usually, the GMAT does consider 0 to be a multiple of every number, right?

Can someone point out my mistake here?

Yes, 0 is a multiple of every number. But we are looking for positive multiples and 0 is neither positive nor negative.
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Why are we not taking a case of a=c?

When a = c, abc − cba = aba − aba = 0 and while 0 IS a multiple of 7, the question specifically asks about the positive multiples of 7 and 0 is not a positive number.
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­(100a+10b+c) - (100c+10b+a) = 99(a-c). This means, a-c is a multiple of 7. To keep the value positive, a must be greater than c. Possible values of (a,c) can be (9,2) and (8,1) -- 2 values. Also, b can take anything between 1 to 9 -- 9 values. Total 2*9=18 values. Option (E) is correct.