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

Good explanation, exactly how I solved it. I love questions with elegant solutions like this. +1

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!

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.

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.

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".

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!

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

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)

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?

(100a + 10b +c) + (100a + b + 10c) + (100b + 10a + c) + (100b + a + 10 c) + (100c + 10b +a) + (100c + b + 10c)
= (200a + 200b + 200c) + (20a + 20b + 20c) + 2a +2b +2c = (a+b+c)(200+20+2) = 222*(a+b+c)

So, I think E.

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