cypherskull wrote:A six digit number is formed by using digits 1,2,3,4,5 and 6 without repeating any of them. What is the sum of all such possible numbers.

a. 279999720
b. 625478924
c. 145269875
d. 149625785

Only 4 answer choices, so clearly this isn't a real GMAT question - please always post your source we know if it's relevant to the GMAT!

This question is a far more complicated version of a question that may appear on the GMAT, so it's worth discussing. Once you understand how one of these questions work, you should be able to tackle any variation.

The key is to first realize that there's no way we're expected to calculate such a monster number, so there must be a trick. Since when do addition problems we add from right to left (i.e. we start with the units digits, then the tens digits, etc...), all we need to do is add enough columns to eliminate all but 1 choice.

First, let's calculate the number of terms in our sum. Since we can't repeat any digits, there will be 6! different ways to arrange the 6 digits that we have.

6! = 6*5*4*3*2*1 = 720

Since we're using each digit an equal number of times (i.e. we'll have just as many numbers that end in 1 as in 2, 3, 4, 5 or 6), there will be 720/6 = 120 of each number in the units column.

**Reward for paying attention time!!**

The next step is to add the numbers 1 through 6 and then multiply by 120 to find the units digit. HOWEVER, no matter what they add up to, when we multiply by 120 we'll get a "0" as our last digit: choose (A), the only answer ending in 0.

If there had been more than 1 answer ending in 0, we'd have to do a bit more work.

1+2+3+4+5+6 = 21
120 * 21 = 2520

Assuming that there were more than 1 answer ending in 0, we'd now have to calculate the second last digit.

Last digit: 0, carry the 252
Second last digit: 2520 + the 252 that we carried = 2772... 2, carry the 277.

We now know that our last two digits are "20" - almost certainly enough info to narrow it down to just 1 choice.

As far as actual GMAT questions go, I've never seen a version of this one with more than 4 digits.

cypherskull wrote:A six digit number is formed by using digits 1,2,3,4,5 and 6 without repeating any of them. What is the sum of all such possible numbers.

a. 279999720
b. 625478924
c. 145269875
d. 149625785

Certainly not a GMAT question.

Please do not worry users by posting such questions!

the answer is a.

6 factorial = 720... 720/6 digits =120 times the number is repeated... 120 x 21 where 21 is 6+5+4+3+2+1 so the total of all 4 digit numbers formed is 2520(111111)=279999720...

any idea how to do it with repeats?

it is clear now that we have total 720 combinations
now it is also understood that every number will be repeated 120 times on each position {Unit place (A),Ten (B),Hundred(C),Thousand(D), Ten Thousand(E), Lacs(F)}
for each row total will be the addition of (1x120) + (2x120) + (3x120) + (4x120) + (5x120) + (6x120)
= 120+240+360+480+600+720
=2520
so on units place: we got "0"(A) and 252 as carry
for Tens place: we get 2520 + 252 (carry)= 2772; "2"(B) fixed for Tens place and 277 is carry
for Hundred Place: We get 2520 + 277 (carry)= 2797; "7"(C) fixed for hundred place and 279 is carry
for thousand place: We get 2520 + 279 (carry)= 2799; "7"(D) fixed for thousand place and 279 is carry
for ...
for Lacs place: We get 2520 + 279 (carry)= 2799 (F); last number
the result will be as shown:
(F) (E)(D)(C)(B)(A)
2799 9 9 7 2 0

this can be applied any where in such questions.