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

Question

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I have never really understood the thinking behind this...

Show Spoiler

OA E

(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

Bunuel · Accepted Answer

Using THREE non-zero digits a,b,c only, we can construct 3!=6 numbers: abc, acb, bac, bca, cab, cba. Their sum will be:

x=

=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+ 
 +(100b+10c+a)+(100c+10a+b)+(100c+10b+a)=

=200*(a+b+c)+20*(a+b+c)+2*(a+b+c)=

=222*(a+b+c)

Largest integer by which x  MUST  be divisible is  222 .

Answer: E (222).

walker · Answer

We can also solve this one without math using symmetry: hundreds, tens and units are symmetric, so sum can be written as (y)*111. We need to check that y is even. For example, for fixed a at hundred position, there is two bc,cb combinations. Therefore, a is included twice (even number of times) into sum of hundreds. So, it is 222

By the way, it is the first time when I add something after Bunuel

shameekv · Answer

Hi Bunuel,
Can you please explain me what will be the value of "x" in this question. If it were asked what is the value of x?

Thanks!

Bunuel · Answer

We cannot say what x is.

If a, b, and c, are 1, 2, and 3 respectively, then x = 123 + 132 + 213 + 231 + 312 + 321 = 1,332 = 6*222 (the least possible value of x).
...
If a, b, and c, are 7, 8, and 9 respectively, then x = 789 + 798 + 879 + 897 + 978 + 987 = 5,328 = 24*222 (the greatest possible value of x).

Hope it helps.

pritishpratap · Answer

Shamee, to solve the problem in a simpler manner why don't you assume the numbers a, b and c to be 1, 2 and 3 respectively?

Thus, the distinct numbers that can be formed would be - 
123
132
213
231
312
321

If you sum these up you get a total of 1332.

Then proceed to plug in the answer options to find the greatest number that divides 1332.

From the options - 
(A) 3 - Yes
(B) 6 - Yes 
(C) 11 - No
(D) 22 - No
(E) 222 - Yes

Clearly, since 222 is the greatest, E is the right option.

KarishmaB · Answer

Here is the catch in "assuming values" in this question: 
The question is a "must be true" question. How do you know that what holds for values 1, 2 and 3 will be true for values say 2, 3 and 7 too? What if sum of numbers formed by 2, 3 and 7 is not divisible by 222? You do need to apply logic to confirm "must be true".

CrackverbalGMAT · Answer

In my humble opinion, although this is not exactly a problem that is tested extensively on the GMAT, there’s no harm in solving questions like these to strengthen your concepts of Permutations with numbers.

If a, b and c are the digits from which we have to form 3-digit numbers such that all the digits are non-zero and distinct, then, the number of ways of doing that is given by,

(a+b+c) * (3-1)! * (111).

In general, if there are ‘n’ distinct digits using which we have to form ‘n’ digit numbers where the digits are non-zero and distinct, then the number of ways of doing this is given by, 
(Sum of the n digits) *(n-1)! *(1111….. n times).

The reason for this is as follows:
When a comes in the units place, its place value will be a. There will be (n-1)! numbers where a will be the units digit. Therefore, the sum of these values will be a * (n-1)!

When a comes in the tens place, its value will be 10a. There will be (n-1)! numbers where a will be the tens digit. The sum of these values will be 10a * (n-1)!

When a comes in the hundreds place, its value will be 100a. There will be (n-1)! numbers where a will be the hundreds digit. The sum of these values will be 100a * (n-1)!.

The sum of all these values will be (n-1)! * a (100 + 10 + 1) = (n-1)! * a * 111.

Similarly, for the other digits, b, c, d and so on, the sum of the values can be worked out as shown above.

In our case, since the sum total is (a+b+c) * 2! * 111 which is nothing but (a+b+c) * 222, we can say that the sum will be definitely divisible by 222. 
The correct answer option is E.

Hope this helps!

aritranandy8 · Answer

VeritasKarishma  Can you plz solve this didn't understand Bunuel explanation

KarishmaB · Answer

Bunuel's method is the most straightforward. I am guessing you are having difficulty understanding 100a + 10b + c.

A 3 digit number abc can be written as 100a + 10b + c (because a is in hundreds place, b in tens and c in units place)
Say 362 = 300 + 60 + 2

How would you add a 3 digit number abc with another 3 digits number cab? 
Say add 362 with 236. You add 300 to 200, 60 to 30 and 2 and 6.

Similarly, abc = 100a + 10b + c
cab = 100c + 10a + b

abc + cab = 100a + 10b + c + 100c + 10a + b

So when you add all 6 numbers you can form with a, b, and c, you will get a's in hundreds place twice, b's in hundreds place twice and c's in hundreds place twice. Same thing for tens place and units place.

Sum = 100 (2a + 2b + 2c) + 10(2a + 2b + 2c) + (2a + 2b + 2c)
Sum = (2a + 2b + 2c)(100 + 10 + 1) = (2a + 2b + 2c)*111
Sum = 222 (a + b + c)

TarunKumar1234 · Answer

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I have never really understood the thinking behind this...

Show Spoiler

OA E

utopianbloke · Answer

There's an ambiguity here. They haven't mentioned if it there was a repetition allowed or not. In the case of repetition, this is the process.

We have a formula for  Sum of all the numbers which can be formed by using the n digits (repetition being allowed) is: (n^n−1)∗(sum of the digits)∗(111... n times).

Substituting values, we get (3^2)(a+b+c)(111) = 999*(a+b+c)

The largest integer from options is 11.

Kinshook · Answer

If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
(A) 3
(B) 6
(C) 11
(D) 22
(E) 222

I have never really understood the thinking behind this...

Show Spoiler

OA E

The numbers = {abc, cba, bac, cab, bca, acb}

x = 200(a+b+c) + 20(a+b+c) + 2(a+b+c) = 222(a+b+c)

IMO E

Vibhatu · Answer

sum of the all numbers using 3 digits= (n-1)! (111...n times) (sum of the digits) 
From the question; (3-1)! (111) (a+b+c) = 2 * 111 * (a+b+c) = 222 (a+b+c)
Largest number is 222. 
 Answer: E

Ved22 · Answer

but  karishma , i tried solving this question using the symmetry concept. In which out of 3 digits no, the first number can be chosen in 1 way and the other 2 number by 8 and 7 ways. this result * 9 ( starting digits go from 1-9). somehow couldnt get the answer. pls help on how to solve using the symmetry concept?

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

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