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.

single digit no formed = 2,4,6,8
sum= 2+4+6+8= 20
two digit= 24,26,28,42,46,48,62,64,68,82,84,86
sum= 660
three digit no= there are possiblity of three like all possiblity from 246,268,468,248
sum of all possiblity=(n-1)!*(sum of the n digits)*111)
using the above formula--
sum of 246 possiblity = 2*12*111=2664
sum of 268 possiblity = 2*16*111=3552
sum of 468 possiblity = 2*18*111= 3996
sum of 248 possibility = 2*14*111= 3108

sum of all 4 digit nos.= 6*20*1111=(atleast use that much of brain of urself..:)

total sum= 147320
sum=

single digit(one) sum of digit is=(2+4+6+8)=20-------------(1)
sum of two digit no is = 11(two digit) *( no of times digit repeat )* (sum of all digit)
=11*3*(2+4+6+8)
=11*3*20 =660 --------------(2)
sum of three digit no is =111*6*20
=13320 ----------------------(3)
sum of four digit no is = 1111*6*20
= 133320 --------------------(4)

adding eqn 1,2,3 and 4
we get sum of all no by forming digit (total sum) =147320

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

4+12+24+24=64

single digit no = 4
2 digit no= 12
3 digit no= 24
4 digit no =24

total for the 1 digit no = (2+4+6+8 =20)
total for the 2 digit no => sum of digits the unit's place + sum of digits at ten's place
sum of the digits in the unit's place =>
as each of the 2,4,6,8 will repeat 12/4 =3 times in at units place then sum = (3*20 =60)
similarly sum of the digits at ten's place =60
so total of 2 digit no =60(10^0 +10^1)=660
total of 3 digit no = sum of digits at unit'place + sum of digits at tens place +sum of digits at hndrd's place
sum of digits at unit's place =>
as each of 2,4,6,8 will repeat 24/4=6 times at units place then sum =(6*20=120)
similarly at the tens and hundred place it will repeat 6 times so sum =120
total of 3 digit no =>120*(10^0 +10^1+10^2)=13320
similarly total of 4 digit no =>120*(10^0 +10^1+10^2+10^3) =133300
sum of all no's => 20+660+13320+133320 = 147320

This question can be solved by using a formula.
1. add the given numbers
here, 2+4+6+8= 20
2. Multiply the resultant by no. of 1s each for a digit
for example for 2( one digit)- it will be multiplied by 1
2,4( two digits)- it will be multiplied by 11
2,4,6( three digits) it will be multiplied by 111
2,4,6,8( four digits) it will be multiplied by 1111
3. now, multiply the resultant by no. of times any digit appears in any position.or u can say no. of times a digit will repeats itself at a fixed position
1 digit can be placed in one position in one way only
for 2 digits- there can be 2 positions_ _ if i fix any digit at any place then other positions can be filled in 3 ways.
for 3 digits - there can be 3 positions _ _ _ if i fix any digit at any place then other positions can be filled in 3*2= 6 ways ( as there are 4 digits given )
for four digits there are 4 position _ _ _ _ if i fix any one digit at any one place others can be filled in 3*2*1 =6

now ,
1. for 1 digit like 2= 20*1*1= 20
2. for 2 digits like 24= 20*11*3 =660
3. for 3 digits like 684 = 20*111*6=13320
4. for 4 digits like 4862= 20*1111*6= 133320
so the answer= 133320+13320+660+20= 147320

wrong question

single digit no. will be 4
2 digit no. will be 12
3 digit no. will be 24
4 digit no. will be 24
so total will be 64

ans is 133320.there is a direct formula. (n-1)!*sum of the digits*1111......continue till n terms.it is vaid wen repeatation is not allowed.

its formula is n-1!(sum of digits)*(1111....)ntimes
in this qstn n=4 so n-1 =3
sum of digits = 20



actually the digits are 2,4,6,8 ,, sum 20
lets cum frm thousands place so _ _ _ _ in this let 2 be in thousands place then rest can be arranged in
3! ways.. therefore it is 2000*3!
like wise formula can be 3!(20)(1111)= 133320

4!=24
so
sum of numbers=n*(n+1)/2=24*23/2=276

2+4+6+8=20for single digit numbers
(2+4+6+8)(3p1*11)=660 sum for double digit number
20*3p2*111=13320 sum for triple digit numbers
20*3p3*1111=22220 sum for four digit numbers
so and is 20+660+13320+22220=36220

this is for 4 digit number
6*(2+4+6+8)*1000
6*20*100
6*20*10
6*10*1
total sum =133320
for 3 digit number
6*20*100
6*20*10
6*20*1
total sum=13320
for 2 digit number
6*20*10
6*20*1
total sum=1320
for 1 digit
6*20*1=120
total sum=133320+13320+1320+120=148080(**might have calculation error but method is correct**)

If 4 digit number is given then i have simple trick...
1111 * 6*(2 + 4 + 6 + 8)


suppose problem is given 3 numbers like 2,4 and 8...then formula is :
111 * (the number of times any digit appears in any position) * (2 + 4 + 8)