1Something is strange subho-
your answer is 1599960 and the answer given in my book is 5199960
1 & 5 interchanged!!! pretty interesting....
Find the sum of all numbers greater than 10,000 formed by using the digits 0,2,4,6,8, no digit being repeated in any number.
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10 Answers
49(2,4,6,8) (the remaining dig..) (all except in 1 5 n 4) (0,2,4,6,8 except tht in 1 n last) (0,2,4,6,8 except that in 1)
we first fill in the first blank...
we first use 2....
numbers...
20468
20486
20846
20864
20684
20648
in the above numbers we can observe one thing....
fixing 2 and 0 in the t.th. n th. places,
4,6,8 occurs in ones tens n hunds place twice each
again 0 can take place of any of the 3 nos...
thus, fixing 2 at t.th place we can have 6 0s,6 2s, 6 4s, 6 8s in each of the ones, tens, hundreds, thousands place...which will become obvious with a bit of thinking and a lot of practice...
thus similarly we can have 4,6,8 in ten thousands place..
thus for all,
we will have 6 each of 2s 4s 6s n 8s in ten thousand's place..
again it will be evident that we will have 18 each of 2s 4s 6s n 8s in rest of places and 24 0s(for whch there is no need of calculation..)
thus the sum is...
10000x6(2+4+6+8) + 1000x18(2+4+6+8) + 100x18(2+4+6+8) + 10x18(2+4+6+8) + 18(2+4+6+8)
please verify answer there is a huge chance of overlooking some cases as this i typed in a hurry! so please someone check out for me!
149nah i cannot miss out such a huge number...! :-o
wait let me check once!
49feel so...unless anyone comes up to save us outta this problem to show i am wrong! :D
1I'm there na.... lol
consider 5 boxes of digits
first digit can be filled in 5 ways (including 0)........next 4 ways and so on.....
so we have 5! = 120 numbers in total......
By symmetry there are 120/5 = 24 nos. starting with each digit (again including 0)
If we write all these numbers one below the other.....
then we have 2 occuring 24 times in first column......4 occuring 24 times in first column and so on.....
So total of each column = (0 + 2 + 4 + 6 + 8) x 24 = 480
Hence sum of all numbers = 480 (104 + 103 + 102 + 10 + 1) = 5333280
Now this also includes 3 digit nos. as we took 0 in first digit....so we need to subtract them.....fix 0 in first box and then consider remaining 4 boxes.....which can be filled in 4! ways......again by symmetry we have each digit occuring 4! = 24/4 = 6 times in each column.
So sum of nos. in each column = (2 + 4 + 6 + 8) x 6 =120
Sum on all nos = 120 (103 + 102 + 10 + 1) = 133320
Subtract this from the prev. answer---
net sum = 5199960
30The answer given is correct.
There's a slight error in Subho's solution.
In the 10k place,
Every digit (2,4,6,8) will appear 4! times instead of 6 times. We've neglected those cases where the zero comes in hundredth, tenth and unit places.
So the answer is
(10000x24+1000x18+100x18+10x18+18)(2+4+6+8)
30Except for the ten thousandth place, the digits (2,4,6,8) in all other places will appear 18 times.
For example,
At the thousandth place,
Since zero at the thousandth place doesnt contribute to the sum.
We have four slots.
_ n _ _ _ [where n is 2,4,6,8]
The first slot can be filled in 3 ways.
The third also in 3 ways(2 remaining digits and also 0)
The second in 2 ways and the last in 1 way.
Hence, a digit in the thousandth place appears for 3x3x2x1 or 3.3! times.
Similarly, you can see for other places also.