Latest Elitmus Aptitude Question SOLUTION: How many 4 digits numbers divisible by 5 can be formed with disgits 0,1,2,3,4,5,6 and 6 a) 220 b) 249 c) 432 d) 216

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Permutation and Combination

Elitmus Question Papers:

(#M40038080) ELITMUS Numerical Ability question Permutation And Combination Keep an EYE

How many 4 digits numbers divisible by 5 can be formed with disgits 0,1,2,3,4,5,6 and 6

a) 220
b) 249
c) 432
d) 216

Asked In Elitmus Ram Neek Singh (9 years ago)

Unsolved Read Solution (21)

Is this Puzzle helpful?   (15)   (7) Submit Your Solution Permutation and Combination

Read Solution (21) :

soumil sharma  (9 years ago)

taking one 6 fixed at first place,

6 _ _ 0 this will give, 6*5=30

6 _ _ 5 this will give, 6*5=30

now confusion of double 6 is gone :)

1 _ _ 0 this can give us 21 (66 is there always)
2 _ _ 0 this can give us 21
3 _ _ 0 this can give us 21
4 _ _ 0 this can give us 21
5 _ _ 0 this can give us 21

1 _ _ 5 this can give us 21 (66 is there always)
2 _ _ 5 this can give us 21
3 _ _ 5 this can give us 21
4 _ _ 5 this can give us 21


so TOTAL = 30+30+(21*5)+(21*4)=249

Is this solution Helpfull? Yes (46) | No (8)  

RAJESH YADAV  (9 years ago)

If given digits are 0,1,2,3,4,5 and 6.
Then
A no. divisible by 5 must end with 0 or 5.
i.e. ( _ _ _ 0 ) and ( _ _ _ 5 ) can be the no.
So, ( 6*5*4 + 5*5*4 ) = 220 = Answer.

Is this solution Helpfull? Yes (19) | No (9)  

rakesh kumar  (9 years ago)

4 digit no which is divisible by 5
= 6*6*6*2
=432

Is this solution Helpfull? Yes (10) | No (18)  

Liti  (9 years ago)

repetition is allowed or not ?? else options are wrong..

Is this solution Helpfull? Yes (6) | No (5)  

Suvendu Karmakar  (9 years ago)

@Ram Neek Singh
1_ _ 0 gives 21 as for avoiding the two 6s confusion lets assume its one 6.
So the above case gives 5*4 numbers i.e 20 numbers . Now considering 1 more six which causes the only case 1660 , thus made the count 20+1 =21 . Same for every other number at thousandth place . So ans is 249 .

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

priyank rastogi  (9 years ago)

249 is the correct ans.

Is this solution Helpfull? Yes (5) | No (4)  

Amit kumar jha  (9 years ago)

I think option is wrong ,,,since questn given 2 six,,if in questn contain one six then option might match ,,,,but if given 2 six so it cant be answer

Is this solution Helpfull? Yes (4) | No (8)  

suyash  (9 years ago)

we have 4 places [1],[2],[3],[4]
we know that a no is divisible by 5,if unit place contains either 0 or 5.
ok now we fix 0 at 4t place then we can fill place 1 with 6 digit,place 2 with 5 digit,place 3 with 4=6*5*4*1=120 ways
now we fix no 5 at 4t position then we can fill position 1 with 5 way "(insted of 6 cz we can not place 0 in to 1st place else it'll become 3 digit no).so place 2 with 5 way and plae 3 with 4 way=5*5*4=100 ways.
so total no-120+100=220 Ans

Is this solution Helpfull? Yes (3) | No (9)  

prasanna  (9 years ago)

6*5*4*2=240

Is this solution Helpfull? Yes (2) | No (9)  

Ram Neek Singh  (9 years ago)

Answer is 249. I don't know how but
Question is right.

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

Anurag srivastava  (9 years ago)

A no is divisible by 5 only if it's last place is 0 or 5 so fix those no
total way for last place is 2

---0/5.

now for first place except 0 all the nos may come.

total way is 6.

6*6*6*2=432
is the ans.

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

ritesh kumar  (9 years ago)

249 is correct
assume one 6 then it will give 220
so the ans must be greater than 220
now count manually by putting 2 six together it will give 29
so 220+29=249

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

anan kumar  (9 years ago)

suyash hum aman tera galat hai 249 hi hoga

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

Bhanuprakash  (9 years ago)

The number must be divisible by 5 so last digit must be 0/5 ,so for the 4thdigit there is a possibility of 2 chances and remaining 3digits are 6*6*6=216

Then possibilities are 216*2=432.

And=432

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

Arun S B  (9 years ago)

432 is correct

Is this solution Helpfull? Yes (1) | No   

aryan rx  (8 years ago)

4 digit nmbr _ _ _ _ ,,,,divisible by 5 means last digit should be 0 or 5 means 2 ways(_ _ _ 0/5)
including 0 at first place now total no of way is 6*5*4*2=240
for 0 at first,,no of way is possible (0 _ _ 0/5) means 5*4=20
now 240-20=220
:)

Is this solution Helpfull? Yes (1) | No   

sooraj singh  (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

Is this solution Helpfull? Yes (1) | No   

Amit kumar jha  (9 years ago)

1) place 0 in extreme position filled 3 positrion =5*6*7/2=105
2) place 5 in extreme position filled remainig 3 posituion= 6*6*5/2=90
so totAL =195

Is this solution Helpfull? Yes | No (12)  

Ram Neek Singh  (9 years ago)

Soumil Sharma @Can you explain how you you get 21
1_ _0 gives 21 how?

@priyanka rastogi tell the steps. Its really confusing question

Is this solution Helpfull? Yes | No (1)  

aditya wargis  (9 years ago)

correct option is (d)
bcz here two sixes are given which is same so we divide the no by 2
6*6*6*2/2=216

Is this solution Helpfull? Yes | No (2)  

swetha R  (9 years ago)

how u ppl are getting 6*6*6 for 1st 3places???

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.