Latest Infosys Aptitude Question SOLUTION: Q. How many 4 digit no can be formed if 1)the no will be an even no 2)dey must be divisible by 4 3)no digit is repeated

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(#M40031918) INFOSYS Numerical Ability question Keep an EYE

Q. How many 4 digit no can be formed if
1)the no will be an even no
2)dey must be divisible by 4
3)no digit is repeated

Asked In Infosys Shilpi Jain (10 years ago)

Solved DHAKAD NEETENDRA SINGH Read Solution (12)

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Read Solution (12) :

DHAKAD NEETENDRA SINGH  (10 years ago)

I think all these parts belong to the same question
like=
How many 4 digit no. can be formed if the no must be divisible by 4 no digit is repeated.
solution:
divisibility for 4= last 2 digit divisible by 4
so,
04,08,20,40,60,80 are last digits with zeroes in it
number of such no's= (8*7*1)*6 = 336
and
without zero are 12,16,24,28,32,36,48,52,56,64,68,72,76,84,92,96 (no repeating digits are allowed)

number of such no's=(7*7*1)*16 = 784
now,
total number of no's which are divisible by 4 = 336+784= 1120 ans...

Is this solution Helpfull? Yes (18) | No (3)  

RAKESH  (10 years ago)

a no. divisible by 4 is always even
last two digit should be 04,08,12,16,20,...,96 (except 44,88) (total 22)
04,08,20,40,60,80 (6) => 6*(8*7)=336
12,16,24,...,92,96 (16) => 16*(7*7)=784
total number of 4 digits which divisible by 4 = 336+784= 1120

Is this solution Helpfull? Yes (13) | No (1)  

SUGANDHA SAGAR  (10 years ago)

1) 9*10*10*5=4500
2) ??????? don't know
3) 9*9*8*7=4536

Is this solution Helpfull? Yes (6) | No (16)  

varunjoshi  (10 years ago)

answer is 1120 and how-->as you know if a number is divisible by 4 then last two digit of number must be divisible by 4 which can be 04,08,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96,00 but and condition is reteatition is not allowed so u have to skip 00,44,88 from above list now as all number which are divisible by four is even number so u need not to think over it. now from above list 6 combinations are those where 0 is available now as repeatition is not allowed so at unit place u can put u can use 8 remaining digit and 10th place u can put any of 7 digit which u havent used 8*7*6(as 6 combinations are dere where 0 is available) in same way for those combination where 0 is not dere u can fill unit place by 7 ways(skip 0 and two which are in use at 100 and 1000 place)*7(except those 3 which u have used at unit 100 and 1000 place) so 8*7*16(combinations)--> so finally 6*8*7+16*7*7=1120

Is this solution Helpfull? Yes (6) | No (2)  

Vijay Singh Adhikari  (10 years ago)

1106 is the ans .

Is this solution Helpfull? Yes (3) | No (2)  

Krishna Charan  (10 years ago)

@sugandha sagar
all those belong to the same question

Is this solution Helpfull? Yes (1) | No   

SUGANDHA SAGAR  (10 years ago)

sorry, i thought all 3 ques are different

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

Prarabdh Singh  (10 years ago)

solve this problem through options

Is this solution Helpfull? Yes | No (1)  

Ayush Rastogi  (10 years ago)

this is onlyone question all of the option are its conditions

Is this solution Helpfull? Yes | No (2)  

praveen prakash  (10 years ago)

getting 1106..(if repetation not allowed)
if 0 is last digit...7*8*4=224
if 2 is last digit...7*7*5=245
if 4 is last digit...7*7*4=196
if 6 is last digit...7*7*5=245
if 8 is last digit...7*7*4=196
total=1106

Is this solution Helpfull? Yes | No (1)  

jatin goyal  (9 years ago)

1)9*10*10*5
2)9*10*5*5
3)9*9*8*7

Is this solution Helpfull? Yes | No   

Suraj  (9 years ago)

6*8*7+16*7*7=1120

Is this solution Helpfull? Yes | No   

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