Latest Elitmus Aptitude Question SOLUTION: How many 4 digit numbers can be formed using digits 0,1,2,3,4,5, such that number is divisible by 8.

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(#M40139325) ELITMUS Numerical Ability question Asked on 18th jan exam Keep an EYE

How many 4 digit numbers can be formed using digits 0,1,2,3,4,5, such that number is divisible by 8.

Asked In Elitmus siddharth prakash singh (9 years ago)

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siddharth prakash singh  (9 years ago)

without repeatation:
case-1 when one of last three digit of them having 0
there are 9 such case (024,032,104,120,240,304,320,504,520)
then no. of ways 3*9=27
case-2 when one of last three digit of them not having 0
there are 5 such case(152,312,352,432,512)
then number of ways 2*5=10
hence total such nos are: 27+10=37

correct me if I am wrong.

Is this solution Helpfull? Yes (39) | No (8)  

AMBARISH SINGH  (9 years ago)

if 4 digit no. is divisible by 8 then last 3 digit must be divisible by 8
so there are 14 case(without repeat)
024,032,104,120,152,240,304,312,320,352,432,504,512,520
on the first place of no. is as 5 type(1,2,3,4,5)
so 14*5=70

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aditya sharma  (9 years ago)

ans is 37..

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

siddharth prakash singh  (9 years ago)

It was not given that whether repeatation is allowed or not.
so we will have to find out he no of ways in both cases, and will have to match with the given choices.

without repeatation:

case-1: when one of last three digit of them having 0
there are 9 such case (024,032,104,120,240,304,320,504,520)
then no. of ways 3*9=27
case-2 when one of last three digit of them not having 0
there are 5 such case(152,312,352,432,512)
then number of ways 2*5=10
hence total such nos are: 27+10=37

With repeataion:
last three digit can be : 000, 024, 032, 040, 104, 120,144, 152, 200, 224, 232, 240, 304, 312, 320, 344, 352,400, 424, 432, 440,504, 512, 520, 544, 552.
so there are 26 ways for last three digit.
and thousand place can be filled by any 5 digit (except 0) so there are 5 ways to fill thousand place.
so total no of ways= 5*26 = 130
Hence total 130 four digit nos are there which are divisible by 8 using digits 0,1,2,3,4 and 5.

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rakesh kumar  (8 years ago)

Applying the divisibility rule of 8 we get
Case 1: _ 024............(3 ways)
Case 2: _032............(3 ways)
Case 3: _104............(3 ways)
Case 4: _120............(3 ways)
Case 5: _152............(2 ways)
Case 6: _240............(3 ways)
Case 7: _304............(3 ways)
Case 8: _312............(2 ways)
Case 9: _320............(3 ways)
Case 10: _352............(2 ways)
Case 11: _432............(2 ways)
Case 12: _504............(3 ways)
Case 13: _512............(2 ways)
Case 14: _520............(3 ways)
So total combinations possible=adding above combinations we get= 3*9 + 2*5= 27+10=37

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Kumari Sweta  (9 years ago)

Ans 300 (without repetition)

1st place- 1,2,3,4,5....... 5 ways (zero will not be included)
2nd place- ....................5 ways (zero may be included)
3rd place-......................4 ways
4th place- 0,2,4.............3 ways( since divisible by 8)

so total- 5x5x4x3= 300

Is this solution Helpfull? Yes (1) | No (16)  

soni kumari  (9 years ago)

for no 024 032 104 120 240 304 320 504 520 thousand digit can be arranged in 3! =6 ways so possible no are 9*6=54
and for 152 312 352 432 thousand digit can be arranged in 2! ways as 0 cant be there
possible no are 5*2=10
total possible no. 10+54=64

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

Vijay Pareek  (9 years ago)

20 without repetiotion

Is this solution Helpfull? Yes | No (8)  

samarendra sahoo  (9 years ago)

For a number divisible 8 the last 3-digit of the number must be divisible by 8.So there are 14 such 3-digit can be formed which are divisible by 8:=024,032,104,120,152,240,304,312,320,352,432,504,512,520
So we can fill the 1st place in 5 ways as we cant place 0 ,
So total 4-diget no which are divisible by 8=14*5=70

Is this solution Helpfull? Yes | No (6)  

Pranjal Shrivastava  (8 years ago)

A no. is div by 8 if last three digit will be div be 8.Among the given no. three digit no.s that will be div by 8 will be:024,032,040,104,112,120,144,152,200,224,232,240,304,312,320,344,352,400,416,424,432,440,504,
512,520,544
Total are 26 and first place can have any of 1,2,3,4,5
Thus answer is 26*5=130

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yagyesh  (8 years ago)

ANS 30
@siddharth 304 will not be the case..... digit '6' is required to fulfil that..... All cases
When repetition not allowed
{024,032,104,120,240,320,352,432,504,512,520}

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