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gmatt1476
The difference 942 − 249 is a positive multiple of 7. 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 ?

A. 142
B. 71
C. 99
D. 20
E. 18

PS01661.01

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
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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


gmatt1476
The difference 942 − 249 is a positive multiple of 7. 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 ?

A. 142
B. 71
C. 99
D. 20
E. 18

PS01661.01
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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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gmatt1476
The difference 942 − 249 is a positive multiple of 7. 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 ?

A. 142
B. 71
C. 99
D. 20
E. 18

PS01661.01

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

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.
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Kanika3agg
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
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rocky620
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
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One query, what if the questions says that a,b,c are distinct integers?
VeritasKarishma mam can you please explain.
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Abcse
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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Adarsh_24
KarishmaB Bunuel
Sorry to bug you guys.

But, why wont something like
-249 and -942 work here?

-249 --942 is +ve multiple of 7.

Numbers are non-zero.

Are negative numbers not considered digits?
I read digits are scalar somewhere.
­
A digit is not negative. We have 10 digits from 0 to 9. When we say "a, b, and c are nonzero digits and the three digit number abc..." we mean that abc is a positive 3 digit number with digits a, b and c. To show a negative number, we will need to use -abc.
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sahiti220620
what are the 9 possibilities referred above? bb Bunuel

The “9 possibilities” are simply the choices for the middle digit b.

Once you fix a and c (like a = 8, c = 1 or a = 9, c = 2), the condition abc - cba being a multiple of 7 is already satisfied. The middle digit b can then be any nonzero digit from 1 through 9.

So for each valid (a, c) pair, there are 9 possible values of b. That’s why the total count is 2 * 9 = 18.

1. 812
2. 822
...
9. 892

1. 912
2. 922
...
9. 992
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One question on this - If a, b and c are given as different alphabets, should we consider them as distinct or not?
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Can we solve the problem this way?

942 = 3^2 2^2 2^1

Total Factors = (2+1) * (2+1) * (1+1) = 18
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eduardolarrranaga
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.
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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? :)
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Why are we not taking a case of a=c?
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nikhilvsh
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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gmatt1476
The difference 942 − 249 is a positive multiple of 7. 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 ?

A. 142
B. 71
C. 99
D. 20
E. 18

PS01661.01
­(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.
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