Quant prob,any takers?

[realme]

hi chk out this sum...
what is the sum of all 4 digit numbers(without repetions).. formed by the digits 0,1,2,4?

[RahulSharma]

When 1st digit is 1, 6 numbers are posiible.

6*1000 + 111*12.

Similarly,
6*1000 + 111*10
6*1000 + 111*6.

Total sum is 18000 + 111*24.

[RahulSharma]

I mean 111*28 + 18000 = 21108.

[whocarez]

hmm....
I am having a different answer

1111*6*7 (for all combinations possible without repetations)

however 0 at the begining doesn't make a valid 4 digit no. subtracting the sum of all numbers when 0 is at first position
111*7*2

thus we have
1111*7*6 -111*7*2=45108

[The Biggest Nawab]

45108 is the correct answer........the associated approach is also valid

[RahulSharma]

024, 042, 204, 240, 402, 420 when 1st digit is 1.
014, 041, 104, 140, 401, 410 when 1st digit is 2.
same for 1st digit is 4.

hence 24, 000 more than what I got.
made a cardinal error in calculation.

[koolfriend4u]

could u explain in brief. how 111*7*2 ??

[pearl]

Plz answer this problem:-
What is remainder of (1^1+2^2+3^3+........+50^50)/8?
Crack it!

[whocarez]

You can see that every even no. terms except for 2^2 would be divisible by 8. Now check the odd no. terms
1 gives 1
3 gives 3
5 gives 5 or -3
7 gives 7 or -1
9 gives 1
11 gives 3
13 gives 5 or -3
15 gives 7 or -1
so set of 4 consecutive odd no.s give are divisible by 8.
There are 25 odd no. terms so ur left with
2^2 + 49^49 (sum of rest is already divisible by
4+1=5 (Ans)