Latest Elitmus Aptitude Question SOLUTION: find the sum of the all the number formed by 2,4,6 and 8 without reputation.Number may be of any digit like 2,24,684,4862.

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(#M40025799) ELITMUS Numerical Ability question qunt Keep an EYE

find the sum of the all the number formed by 2,4,6 and 8 without reputation.Number may be of any digit like 2,24,684,4862.

Asked In Elitmus Alok kumar (10 years ago)

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Read Solution (8) :

SOURAV KUNWAR  (10 years ago)

(2+4+8+6)*3!*1111 = 133320 (CORRECT ANSWER)..

Is this solution Helpfull? Yes (17) | No (7)  

Anuratha  (10 years ago)

2+4+6+8=20
24+26+28+46+48+68=240
42+62+82+64+84+86=420
246+264+426+462+624+642=2664
268+286+628+682+826+862=3552
248+284+428+482+824+842=3108
468+486+648+684+846+864=3996
2468+2486+2648+2684+2846+2864=15996
4268+4286+4628+4682+4826+4862=27552
6248+6284+6428+6482+6824+6842=39108
8246+8264+8426+8462+8624+8642=50664
so,20+240+420+2664+3552+3108+3996+15996+27552+39108+50664=147320

Is this solution Helpfull? Yes (15) | No (7)  

rohit raj  (10 years ago)

1 digit no sum=2+4+6+8=20
2 digit no sum=22*3+44*3+66*3+88*3=3*220
3 digit no sum=222*6+444*6+666*6+888*6=6*2220
4 digit no sum=2222*6+4444*6+6666*6+8888*6=6*22220
now add all(22220+2220)*6+660+20=147320
explanation:
for 1 digit it is simple
for 2 digit no we know if we place 2 at 10th position then the other 1 digit may be taken from rest(4,6,8)and if we place 2 at 1th position then also the 10th position filled by one of from(4,6,8). thus 20*3+2*3=22*3 and same thing will happen for 4 6 8 thus for 2 digit sum will be 22*3+44*3+66*3+88*3=3*220
....for 3 digit we can choose 3 out of 4 by 4 ways,no made by one of these groups(246,468,268,248)we can see every digit comes in three groups.we take 2 first for 2 at 100th this thing will happen 2 time in one group that is total 6 time bcoz there are 3 groups simillarly for 10th place and for 1th place thus
200*6+20*6+2*6=222*6 thus total 222*6+444*6+666*6+888*6=6*2220
.....lastly for 4 digit 2468 if we place 2 at any position other can arranged by 3!=6 different way that means 2000*6+200*6+20*6+2*6=2222*6 same for all thus
2222*6+4444*6+6666*6+8888*6=6*22220

Is this solution Helpfull? Yes (15) | No (10)  

Anand Kumar  (10 years ago)

Let,
No of digits : m
No of digits for which we are finding he sum : n
sum of digits : sod

Suppose we are finding all 3 digit no from 4 digits(2,4,6,8):

First we will find for unit place : how many times 2 will come :
2 is fix now, so other arrangement = 3P2 i.e 6 (m-1Pn-1)
similarlly all digits will come at unit place.
similarlly for tens place ...

So Sum of n digits from m digits,S :
= (1*sod* m-1P n-1)+
(10*sod* m-1P n-1) +
(100 * sod * m-1P n-1) +
...
...
(10^n-1 * sod * m-1P n-1)
S(m,n) = sod* m-1 P n-1 * 111...1(n times)

_________________________________________________
puting this in this question :

Sum of 4 digit numbers = (2+4+6+8)* 3P3 * (1111) = 20*6*1111 = 133320
Sum of 3 digit numbers = (2+4+6+8)* 3P2 * (111) = 20*6*111 = 13320
Sum of 2 digit numbers = (2+4+6+8)* 3P1 * (11) = 20*3*11 = 660
Sum of 1 digit numbers = (2+4+6+8)* 3P0 * (1) = 20*1*1 = 20

Adding All ,
Sum = 147320

Is this solution Helpfull? Yes (13) | No (3)  

Vinay Gupta  (10 years ago)

there is a formula to solve such type of problem :
(sum of all digits) * 1111...(r times) * (n-1)P(r-1)

where
P - Permutation
n - Total given digits
r - No. of digits in a number

So solution like as following:

1. for 2+4+6+8 => n=4,r=1 -> (2+4+6+8)*1*(4-1)P(1-1) = 20

2. for 24+26+28+46+48+68+42+62+82+64+84+86 => n=4,r=2 -> (2+4+6+8)*11*(4-1)P(2-1) = 660

3. similarly n=4,r=3 -> (2+4+6+8)*111*(4-1)P(3-1) = 13320

4. and n=4,r=4 -> (2+4+6+8)*1111*(4-1)P(4-1) = 133320

So to

Is this solution Helpfull? Yes (10) | No (1)  

Prerna  (10 years ago)

143720 will be sum of all the numbers formed by 2,4,6 and 8 without reputation.

Is this solution Helpfull? Yes (2) | No   

Alok kumar  (10 years ago)

is there any other method to solve this q???

Is this solution Helpfull? Yes (1) | No (1)  

vinod  (10 years ago)

147320 is correct

Is this solution Helpfull? Yes (1) | No   

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