Latest Elitmus Aptitude Question SOLUTION: How many numbers can be fromed from the digits 0,2,4,6,8 which are divisible by 8

I think 30 numbers if only 5 digit numbers are allowed and all 5 given digits are to be used once only.

5 dig no.-30,4dig no.-30,3dig no.- 15, 2 dig no.- 5, 1dig no.- 2
as no. wiht last 3 dig div by 8 is div by 5 hence total no.=60+60+15+7=82

considering numbers are not repeated.

watevr first 3 digits are. only that counts is if last 2 digits is divided by 8.

_ _ _ 0 8
_ _ _ 2 4
_ _ _ 4 0
_ _ _ 4 8
_ _ _ 6 4
_ _ _ 8 0
this are 6 variations it can have.

_ _ _ can be arranged in 6 ways.

therefore 6 x 6 =36

@Zen
u have to subtract 6 from 36
ans is 36-6=30
because u havent subtracted the numbers starting from 0

Is repetition of numbers is allowed ?
Is there any limit on number of digits ?

repetion of digits are not allowed
no limit in number of digits

Nnnñdpydxfuiyfiygoy

We need to consider numbers with 1 digit, 2 digit, 3 digit, 4 digit and 5 digit

Case1 : 5 digit numbers

For a number to be divisible by 8, its last three digits must divisible by 8.

So, last three digits can be from the following group
(024), (048), (064), (208), (240), (248), (264), (280), (408), (480), (608), (624), (640), (648), (680), (824), (840), (864),

There are 18 such groups.

Consider these 12 groups in which zero is there
(024), (048), (064), (208), (240), (280), (408), (480), (608), (640), (680), (840)

With each of these 12 groups, first digit can be any of the remaining 2 digits and second digit can be the remaining one digit
Therefore, 12×2 = 24 such numbers

Consider the 6 groups in which zero does not come
(248), (264), (624), (648), (824), (864),

With each of these 6 groups, first digit can be the one remaining digit(because zero cannot be the first digit)
Second digit can be zero
Therefore, 6×1 = 6 such numbers

Therefore, total count of 5 digit numbers = 24 + 6 = 30

Case 2 : 4 digit numbers

We can use similar approach. Consider these 12 groups in which zero is there
(024), (048), (064), (208), (240), (280), (408), (480), (608), (640), (680), (840)
First digit can be any of the remaining two digits
Total count of such numbers = 12×2 = 24

Consider the 6 groups in which zero is not present
(248), (264), (624), (648), (824), (864),
First digit can be only the one remaining non-zero digit
Total count of such numbers = 6×1 = 6

i.e., total count of 4 digit numbers = 24 + 6 = 30

Case 3 : 3 digit numbers

We have already seen that last three digits must be from the following 18 groups
(024), (048), (064), (208), (240), (248), (264), (280), (408), (480), (608), (624), (640), (648), (680), (824), (840), (864)

Therefore, we have 15 valid 3 digit numbers as given below
(208), (240), (248), (264), (280), (408), (480), (608), (624), (640), (648), (680), (824), (840), (864),

i.e., total count of 3 digit numbers = 15

Case 4 : 2 digit numbers
Only 5 numbers (24,40,48,64,80)

Case 5 : 1 digit numbers
2 numbers (0, 8)

Therefore, total count of numbers that can be formed from the digits 0,2,4,6,8 which are divisible by 8
= 30 + 30 + 15 + 5 + 2
= 82

8,24,48,64 (TOTAL IS 4)
-24 IS 2WAYS
-48
-64....... SO TOTAL 6 THREE DIGITS NUMBERS
--24
--48
--64..... SO TOTAL 12 FOUR DIGIT NUMBERS
---24(---48 ----64) SO TOTAL 12 5 DIGIT NUMBERS 1+3+6+12+12=34