Latest TCS Aptitude Question SOLUTION: How many 8 digits number can be formed with repitition by using 1,2,3,4,5 which are divisible by 4 ?

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How many 8 digits number can be formed with repitition by using 1,2,3,4,5 which are divisible by 4 ?

Asked In TCS VIKASH KUMAR SHARMA (10 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

Sourav Malhotra  (10 years ago)

5^7 as we can fill the first 6 digits in 5^6 ways with repetiton and then for the last two digits we have only 5 combinations with the given numbers that is 12,24,32,44 and 52 so we get 5^7 combinations

Is this solution Helpfull? Yes (31) | No   

amit sen  (10 years ago)

sorry....5 possibility r there for filling last two digit(12,24,32,44,52)
sol=5^6*5=78125

Is this solution Helpfull? Yes (11) | No (2)  

Rituparna Bhattacharyya  (10 years ago)

sorry missed 52..the ans 5^7

Is this solution Helpfull? Yes (7) | No (1)  

Rituparna Bhattacharyya  (10 years ago)

ans. 62500
a no to be divisible by 4 should have the last two digits divisible by 4
there can be 4 such no(24,44,32,12)
rest 6 place can be filled by 5 ways.
thus soln = 5^6*4 = 62500

Is this solution Helpfull? Yes (4) | No (10)  

Vishal Sagar  (10 years ago)

the first 6 places can be filled with 5^6,we have to check for the last two digits
for divisibilty by 4 they must be divisible hence at the 8th place we can have only 2 & 4 hence
for 2(as 8th digit)
(5^6)*3*1(as the 7th place can be filled with 1,3,5 only)
for 4(as 8th digit)
(5^6)*2*1(as the 7th place can be filled with 2,4 only)
hence (5^6*3*1)+(5^6*2*1)=78125

Is this solution Helpfull? Yes (3) | No   

vatan singh  (10 years ago)

suppose there r 8 places to be filled by given no. with repition
for no to be divisible by 4 it must contain 12, 24, 32, 44, 52 at last i.e. if last two digits r divisible by 4 then no. is divisible by 4
possible no. r now 6 places left suppose solve for no. contain 12 at last= 5^6 i.e repition allowed but here total possible no. possible other than 12 including 12 at last r= 5^6 * 5= 5^7 is answer

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tauseef arbab  (10 years ago)

answer is 5^7

Is this solution Helpfull? Yes (2) | No   

amit sen  (10 years ago)

A no is divisible by 4 if last two digits r divisible by 4
so,here total 4 possibility(12,24,32,44)
so,last two digits r collectively filled by 4 ways....
rest 6 places r filled by 5 ways each(becaus repeation is allowed)
so,solution= 5^6*4=62500

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

shubhankar barman  (10 years ago)

A no is divisible by 4 if last two digits r divisible by 4
so,here total 4 possibility(12,24,32,44)
so,last two digits r collectively filled by 4 ways;
repeation is allowed so
rest 6 places r filled by 5 ways each
so,solution= 5^6*4=62500

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

prince  (10 years ago)

rite ans ...........5^7.

Is this solution Helpfull? Yes (1) | No   

praveen shahi  (10 years ago)

the answer is 375000.
for the first six places there are . 5 options for each place and for the alst two there are 24 options( nly 24 no. from 1-99 which are divisible by 4 ).
so total no. of permutations = (5^6)*24 = 375000

Is this solution Helpfull? Yes | No (7)  

Venkateswar  (10 years ago)

there are 5 numbers 1,2,3,4,5
we required 8 digit number
so
5^8-1=5^7

Is this solution Helpfull? Yes | No (1)  

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