If S represents the sum of all the positive three-digit numbers that c

Question

If S represents the sum of all the positive three-digit numbers that can be formed using each of the distinct nonzero digits x, y and z exactly once in each integer, which of the following must be a factor of S?

A. 4
B. 5
C. 7
D. 11
E. 111

A. 4
B. 5
C. 7
D. 11
E. 111

gmatophobia · Accepted Answer

Let's take a small example to understand the question -

We know S represents the sum of all positive 3 digit number that can be formed using each of the distinct non-zero digits.

So, x , y and z can take values any values between 1 and 9, both inclusive.

For the sake of simplicity let's take the value of x = 1, y = 2 and z = 3

Possible arrangements are

1 2 3 
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1

Let S be the sum of all these numbers

We can write the numbers as

1 2 3 = 100 + 20 + 3
1 3 2 = 100 + 30 + 2
2 1 3 = 200 + 10 + 3
2 3 1 = 200 + 30 + 1
3 1 2 = 300 + 10 + 2
3 2 1 = 300 + 20 + 1
-------------------------
 S = 12*100 + 12*10 + 12
 S = 12 (100+10+1) = 12 * 111

Hence we see that 111 is a factor of the sum.

Now 12 is nothing but twice the sum of the digits that we have taken.

12 = 2 * (1+2+3)

If we were to take all the digits (1 to 9) the resultant sum can also be represented the form of  x * 111 . The value of x will depend on the digits that we're taking but the form will not change.

Hence 111 must be a factor of S

Option E

Akshaynandurkar · Answer

i did not get the solution. Could you please explain how you have extrapolated the results from 6 numbers to summation of all numbers.

gmatophobia · Answer

Hey  Akshaynandurkar

The numbers selected doesn't matter.

The three-digit number can be formed using 1, 2, 3, ...8, 9. Out of these 9 digits, we need to select three digits.

Assume that we have selected a,b,c. We can create six possible combinations using a,b,c as shown below -

a b c
a c b
b a c
b c a
c a b
c b a

Observe → 'a' repeats twice in the units place, 'b' repeats twice in the units place and 'c' repeats twice in the units place

A similar observation can be made for other places as well (ex. tens place and the hundreds place)

Hence, the sum of all possible combination of abc will be (2*a + 2*b + 2*c)*(100) + (2*a + 2*b + 2*c)*(10) + (2*a + 2*b + 2*c)*(1) = (2a + 2b + 2c)*(111)

Now say we select a different combination of digits, say pqr.

The sum of all possible combination of pqr will be (2p + 2q + 2r)*(111)

Similarly you can select any combination. The sum of all combinations will be

{ (2a + 2b + 2c) + (2p + 2q + 2r) + ( .... ) + ... + ( .... ) } * 111

Hope this helps.

Akshaynandurkar · Answer

Got it. Thank you for the clarification.
Similar way for 4 digit number (without using 0 as a digit), we will have the factor 1111. for 5 digits .....11111
Isn't it ?

If S represents the sum of all the positive three-digit numbers that c

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If S represents the sum of all the positive three-digit numbers that c

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