Latest Elitmus Aptitude Question SOLUTION: how many 4 digit no. divisible by 5 can be formed with the digits 0,1,2,3,4,5,6 and 6.. option 1) 432 2) 249 3) 220 4) 288

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(#M40030565) ELITMUS Numerical Ability question apti Keep an EYE

how many 4 digit no. divisible by 5 can be formed with the digits 0,1,2,3,4,5,6 and 6..
option
1) 432
2) 249
3) 220
4) 288

Asked In Elitmus ankit chauhan (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

Arpit Agarwal  (10 years ago)

it cannot be other than 249
0, 1, 2, 3, 4, 5, 6, 6
case 1 : __ __ __ 0
all are different(1,2,3,4,5,6) = 6C3 * 3! = 120
1 pair of 6 and 1 other number from (1,2,3,4,5) = 5C1 * 3!/2 = 15

case 2 : __ __ __ 5
1 pair of 6 and 1 other number from (1,2,3,4)(not 0) = 4C1 * 3!/2 = 12
1 pair of 6 and 1 zero(0) = 2 (0 cannot be at the first place, u need 4 digit number)
all are different(0,1,2,3,4,6) = 1 place(1,2,3,4,6); 2,3 place(0,1,2,3,4,6) = 5*5*4 =100

= 120+15+12+2+100 = 249

Is this solution Helpfull? Yes (22) | No (2)  

rashmi gitanjali  (10 years ago)

588 with repetition(6*7*7*2)
220 without repetition((6*5*4*1)+(5*5*4*1))

Is this solution Helpfull? Yes (20) | No (8)  

ankit sachan  (10 years ago)

ans is 220

Is this solution Helpfull? Yes (16) | No (5)  

vinod  (10 years ago)

The answer is 249
first of all place the non repeated digits ending with 0 so we get 6p3=120
by placing last with 5 we get 1000 place can be fill with 5 possibles and remaing with 5p2 so we get =5*5p2=100


upto know we get values having without reputation of 6
Now we can go with reputation of 6 ........
first place the unit digit with the 0,we place 2 sixes and one is different and this can be arranged in slecting the different of remaing is 5c1*3!/2!==15
as because six is repeated 2 times then we are going to divide with the 2!


by placing 5 at last we have two posibilities
first place 6 at first then another 6 in other , then other value from 5c1*2!=10
here 2! because 6 and different number can be arranged in 2! ways...
last thing not starting with 6 by placing the last digit value 5 is

first place can we fill in 4c1 ways and remaing two with 6 so the posibilities are 4c1*1=4


By adding all the total we get 249.

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

Dheeraj  (10 years ago)

Without repetition of 6:
Ending with 5 is 100 nos
Ending with 0 is 120 nos

With repetition of 6:
Ending with 5 is 14 nos
Ending with 0 is 15 nos

Total nos are: 249

Is this solution Helpfull? Yes (8) | No (2)  

m.sindhuri  (10 years ago)

answer is 432..
reason:- Since in order to get divisible by 5 ,the units place must be vth 0 or 5 so it is 2! and d remaining nos r 6 and in tens plac dey can b arranged in 6! and hundreds plc 5! and thousands plc 4!=4!+5!+6!+2!=865/2=432.. since 6 is twice.

Is this solution Helpfull? Yes (7) | No (5)  

MD SHOAIB  (10 years ago)

ans is 249

Is this solution Helpfull? Yes (5) | No (1)  

Gaurav  (10 years ago)

6*7*7*2 = 588 Ans.

Is this solution Helpfull? Yes (1) | No (8)  

ajay negi  (10 years ago)

koi shi solution btao iska??

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kundan prasad  (8 years ago)

Ans;3) 220
Sol ; Case I, when the last digit is fix 0 (zero)
5*5*4*1=100
Case II,when the last digit is fix 5
6*5*4*1=120
total 4 digit no divisible by 5 is
Case I+Case II=100+120=220

Is this solution Helpfull? Yes (1) | No (3)  

saloni garg  (10 years ago)

I think answer should be 896

Is this solution Helpfull? Yes | No (7)  

shivendu kumar  (10 years ago)

432.
On unit place we can put (0,5).So we can place on unit place by 2 ways,and other three places we can submit by 6 ways,because repeatation is alowed.
Ans: 6*6*6*2=432

Is this solution Helpfull? Yes | No (6)  

anubhav tyagi  (7 years ago)

arun sharma's permutation and combination lod 2 2nd question

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Damini singhal  (7 years ago)

Unit digit can be chose in 2 ways- either 0 or 5.

When unit digit is 5:
With just a single 6:

Number of numbers possible = 5 * 5 * 4 = 100 (as MSD cannot be a 0)

Now, two 6's can come as 66x 6x6 or x66, and this counts to 5 + 5 + 4 = 14 numbers (for the last case, x can only be from 1-4 and not 0)

So, total numbers ending in 5 = 100 + 14 = 114.

When unit digit is 0:
With just a single 6:

Number of numbers possible = 6 * 5 * 4 = 120

Now, two 6's can come as 66x 6x6 or x66, and this counts to 5 + 5 + 5 = 15 numbers.

So, total numbers ending in 0 = 120 + 15 = 135.

Thus, total numbers divisible by 6 = 114 + 135 = 249.

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Tejasvi Kumar  (6 years ago)

The remainder we get when we divide the sum of the digits by 9 is the same as the remainder when we divide the number by 9
so rem of 22*22/9 = rem of 4*4/9
rem of 222*222/9 = rem of 6*6/9

rem of 2^2+ 4^2+6^2.............98^2/9
rem of 2^2(1+2^2+3^2............+49^2)/9

1+2^2+3^2............+49^2=49(49+1)(98+1)/6=40,425

rem of 4*40425/9 = 161700/9= 1+6+1+7=15
Rem=1+5=6

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