Latest Elitmus Aptitude Question SOLUTION: How many 5 digit numbers can be formed by using 0,2,4,6,8 which is divisible by 8? (for a no. to be divisible by 8, last three digits of the number should be divisible by 8) option

a.) 30
sol url---> http://postimg.org/image/dlavp2ee1/

hey Naveen Kumar FORM WHICH BOOK R PREPARING YAAR .
I HAVE ELITMUS EXAM ON 3 JAN 2016 PLEASE SUGGEST HOW TO PREPARE FOR THAT ..
PLEASE BROOOOOO

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

@ MD IQBAL SIDDIQUI .......... solved previous Elitmus Q. from M4math conceptually i.e enough ...
if ur Cryptarithmetic is fine thats enough for elitmus otherswise follow http://www.elitmuszone.com/
for Crypt & Data sufficiency

1st place can be filled in 4 no of ways (except 0)
2nd place can be filled in 3 no of ways
last 3 places can be filled in 18 no of ways
total 4*3+18=30

(c)26 is answer.

hai naveen can u clearly explain cryptographic question for me please

30 is the r8 answer.

@ VADAKATTU JAGADEESH ..i mentioned the solution URL in Cryp. Q

fix last digit __ __ ___ __ 0
then we can fill tenth digit with 2,4,6, and 8
let put 2 at tenth digit __ ___ __ 20
but then putting Any remaining number at 100th digit we can not get number divisible by 8
So put 4 at tenth digit _ _ _ 40
then if we fill 100th digit with any number we can get number divisible by 8
So total number is =6
So put 8 at tenth digit _ _ _ 80
then if we fill 100th digit with any number we can get number divisible by 8
So total number is =6


same fix last digit __ __ ___ __ 2
then we can fill tenth digit with 0,4,6, and 8
but putting any number we cannot get number divisible by 8


fix last digit __ __ ___ __ 4
then we can fill tenth digit with 2,0,6, and 8
let put 0 at tenth digit __ ___ __ 4
but then putting Any remaining number at 100th digit we can not get number divisible by 8
So put 2 at tenth digit _ _ _ 24
then if we fill 100th digit with any number we can get number divisible by 8
So total number is =6
but two number in which number start with 0
06824 and 08624
So remaining number =4
So put 6 at tenth digit _ _ _ 64
then if we fill 100th digit with any number we can get number divisible by 8
So total number is =6
but we get two number start with 0
02864 and 08264
So remaining number=4

same fix last digit __ __ ___ __ 6
then we can fill tenth digit with 0,4,2, and 8
but putting any number we cannot get number divisible by 8



fix last digit __ __ ___ __ 8
then we can fill tenth digit with 2,0,6, and 4
let put 0 at tenth digit __ ___ __ 8
but then putting Any remaining number at 100th digit we get number divisible by 8
So total number=6
(0 already in used)


So put 4 at tenth digit _ _ _ 48
then if we fill 100th digit with any number we can get number divisible by 8
So total number is =6
but we get two number start with 0
06248 and 02648
So remaining number=4

Answer=6+6+4+4+6+4=30