Latest Elitmus Aptitude Question SOLUTION: que:Find the sum of all the nos which can be formed by 2,4,6,8(repetitions not allowed) ex:2+4+6+8+....24+.....68+.....468+

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(#M40040181) ELITMUS Numerical Ability question permutation and combination Keep an EYE

que:Find the sum of all the nos which can be formed by 2,4,6,8(repetitions not allowed)
ex:2+4+6+8+....24+.....68+.....468+

Asked In Elitmus dharmendra kumar (9 years ago)

Unsolved Read Solution (14)

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SHANTANU RAINA   (9 years ago)

for single digit no arranggement sum =20

total 2 digit no possible =12 and sum is 12/2*(24+86)=66
total for 3 digit no possible=4*3*2 =24 and sum is 24/2*(246+864)=13320
total for 4 digit no possible=4*3*2*1=24 and sum is 24/2(2468+8642)=133320

hence sum is 20+660+13320+133320= 147320 ans.

Is this solution Helpfull? Yes (44) | No (5)  

mahul  (9 years ago)

EX=4*3*2*1
24/4=6(4 is total number)
.>
6*(2+4+6+8)=120,
120*1+120*10+120*100+120*1000,
120*(1111)=133320

Is this solution Helpfull? Yes (12) | No (9)  

dhananjay kumar  (9 years ago)

20+660+13320+133320=147320 ans

Is this solution Helpfull? Yes (12) | No (2)  

Ved Prakash Yadav  (9 years ago)

formula-
(n-1)!(sum of digit)*(1111)
(3)!*20*1111
=133320

Is this solution Helpfull? Yes (10) | No   

Harmanpreet singh  (9 years ago)

2 + 8642 = 8644/2 = 4322

4322 * 64 =276608 is ans.

Is this solution Helpfull? Yes (2) | No (1)  

Harmanpreet singh  (9 years ago)

64 no. can be formed using 2,4,6,8

take min no. that is 2.. and largest no. that is 8642 and then make average of this that is 4322 and multiply it with 64.. u will find the sum of the numbers

Is this solution Helpfull? Yes (2) | No (1)  

Kumar Vikash  (9 years ago)

ans is 147320

Is this solution Helpfull? Yes (2) | No (1)  

ashish ranka  (9 years ago)

kumar vikash...your ans is correct....it was asked on 28 sept...but will u tell me plzz in which option was right....it is option (a).....actually i forgot on which option 147320 was written...is it option a or option b....my set was D ...plz reply...

Is this solution Helpfull? Yes (2) | No   

rochit ranjan  (9 years ago)

total number of single digit number = 4, their sum = (2 + 4 + 6 + 8) = 20
total number of two digit number = 12 and sum of the digits at units place (2 * 3 + 4 * 3 + 6 * 3 + 8 * 3) = 60
sum of two digits number = 60 + 60*10 = 660
total number of three digits number = 4 * 3 * 2 = 24
sum of digits at units place in all the three digits number = (2 + 4 + 6 + 8) * 6 = 120
sum of the three digits numbers = 120 + 120 * 10 + 120 * 100 = 13320
total 4 digit number = 24
sum of all the 4 digits number = 13320 + 120 * 1000
total sum = 147320

Is this solution Helpfull? Yes (2) | No   

dharmendra kumar  (9 years ago)

can u please explain the exact logical reason???@harmanpreet singh

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kalyani kumari sharma  (9 years ago)

sum=s1=(2+4+6+8+24+............+8462) total 64 terms
s2=(8642+8462+..................+2) total 64 terms
s1+s2=(8642+2)*64
sum=s1=s2=(s1+s2)/2=276608

Is this solution Helpfull? Yes | No (2)  

raj  (9 years ago)

For 1 digit max number= 8 and min 2 and avg= 5 * 4(4 is the total number of 1 digit )= 20.
For 2 digit min 28 max 82 avg =50 total number of 2 digit= 12
So 50*12=600
For 3 max 862 min 246 avg 500 * 24 ( 24 is total number of 3 digit
= 12000
And for 4 digit it will be 120000 now add 120000+12000+600+20

Is this solution Helpfull? Yes | No (1)  

dharmendra kumar  (9 years ago)

kumar vikash how did u solve the que???

Is this solution Helpfull? Yes | No (1)  

abhishek maini  (9 years ago)

total number of 1 digit number can be=2,4,6,8 ie 4
total number of 2 digit number can be(repetition not allowed)=4*3=12 ways
total number of 3 digit number can be(repetition not allowed)=4*3*2=24 ways
total number of 4 digit number can be(repetition not allowed)=4*3*2*1=24 ways
now smallest number can be 2
largest number can be 8642
so average of smallest and largest will be=8642+2/2=4322
total selections of 1,2,3,4 digit combinations=4+12+24+24=64
so sum wil be=8644*64=276608

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