Elitmus
Exam
Numerical Ability
Age Problem
form all different number using 2,4,6,8 digits( eg: 2,46,8624) without repetition
find the sum of these numbers.
Read Solution (Total 11)
-
- 4 digits (2,4,6,8) so sum is
(4-1)factorial*(2+4+6+8)*1111 = 133320 - 9 years agoHelpfull: Yes(15) No(2)
- 2+4+6+8=20
24+26+28+42+46+48+62+64+68+82+84+86=660
246+248+264......+864=13320
2468+2486+2648+2684+.....=133320
total: 147320 - 9 years agoHelpfull: Yes(14) No(13)
- (2+4+6+8)(1111)*3!+[(2+4+6)+(4+6+8)+(2+6+8)](111)*2!+[(2+4)+(4+6)+(2+6)+(2+8)](11)+2+4+6+8
ans is 133320+13320+660+20= 147320 - 9 years agoHelpfull: Yes(5) No(2)
- The required numbers are as follows;
8642
8624
8462
8426
8246
8264
6842
6824
6482
6428
6248
6284
4682
4628
4862
4826
4286
4268
2648
2684
2468
2486
2846
2864
sum 133320 - 9 years agoHelpfull: Yes(2) No(1)
- (2+4+6+8)(1111)*3+[(2+4+6)+(4+6+8)+(2+6+8)](111)*2+[(2+4)+(4+6)+(2+6)+(2+8)](11)+2+4+6+8
==666660+10212+374+20=77266 - 9 years agoHelpfull: Yes(1) No(4)
- First condition for single digits = 4
Second for two digit = 4*3 = 12
Third for three = 4*3*2 = 24
Fourth for four digit = 4*3*2*1 = 24
total no. will be = 24+24+12+4 = 64 - 9 years agoHelpfull: Yes(0) No(6)
- what is the answer pls explain
- 9 years agoHelpfull: Yes(0) No(0)
- 4 digit number without repetition
so it will be 4*3*2*1=24 - 9 years agoHelpfull: Yes(0) No(0)
- 13320 is an answer
- 9 years agoHelpfull: Yes(0) No(0)
- The answer is 133320+124432+660+20= 268432
- 9 years agoHelpfull: Yes(0) No(0)
- same solution as abhishek singh .
for single digit sum is = 20
so on. - 9 years agoHelpfull: Yes(0) No(0)
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