Latest Elitmus Aptitude Question SOLUTION: how many 4 digit number can be formed by 0,1,2,3,4,5, number is divisible by 4

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(#M40103700) ELITMUS Numerical Ability question Math Keep an EYE

how many 4 digit number can be formed by 0,1,2,3,4,5, number is divisible by 4

Asked In Elitmus binesh kumar (9 years ago)

Unsolved Read Solution (23)

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Read Solution (23) :

Ravindra Pratap Singh  (9 years ago)

Last two digits can be 00,04,12,20,24,32,40,44,52 that is 9 possibilities for last two digits.For the hundredth place digit all 6 possibilities exist.For the thousand place we have 5 options (1,2,3,4,5 ). Hence the solution is 5*6*9=270

Is this solution Helpfull? Yes (33) | No (14)  

siddharth prakash singh  (9 years ago)

without repeatation:
case-1 when one of last two digit of them having 0 (04,20.40,)then no. of ways 4*3*3=36
case-2 when one of last two digit of them not having 0 (12,24,32,52) then number of ways 3*3*4=36
total no.of ways is 36+36=72 nos.

Is this solution Helpfull? Yes (30) | No (3)  

samarendra sahoo  (9 years ago)

without repetition:

case-1:last 2-digit contains zero-04,20,40(3 ways)so the 100th and 1000th place can be filled up in 4*3 ways.So totla=4*3*3=36 ways
case-2:when last 2-digit doesn't contain zero-12,24,32,52(4 ways) and the 1000th place can be filled up in 3 ways(as we can't place 0 there) and 100th place in 3 ways,So total=3*3*4=36
So total numbers without repetion=36+36=72

With repetition:

00,04,12,20,24,32,40,44,52(9 ways),1000th place in 5 ways(as we can't use 0 ) and 100th place in 6 ways.So total =9*5*6=270

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binesh kumar  (9 years ago)

total ans 144

Is this solution Helpfull? Yes (7) | No (10)  

NESAB JAFAR  (9 years ago)

case-1 when iast two digit one of them having 0(04,20.40,)then no. of ways 4*3*3=36
case-2 when last two digit is 12,32,52,24 then number of ways 4*3*4=48
total no.of ways is 36+48=84

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RAVI KIRAN  (9 years ago)

04,12,20,24,32,40,52 ...... 5*6*7

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ASHISH MUKHERJEE  (9 years ago)

case 1: 4*3*2*1 = 24 (when in unit place we put zero(0))
case 2: 3*3*3*1 = 27 (when in unit place we put two(2))
case 3: 4*3*1*1 = 12 (when in tens place we put zero(0))
case 4: 3*3*1*1 = 9 (when in tens place we put two(2))

so total possible number can be formed = 24+27+21 =72.

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

Abhishek  (9 years ago)

To be divisible by 4 last 2 digit must be divisible by 4. So, the last two digit can be {(0,4), (2,0), (4, 0)} and {(1, 2), (2, 4), (3, 2), (5,2)}
So, there are three ways where last two digit have 0 and there are 4 ways where last two digit are non-zero. The remaining 2 digits can be arranged in following way
Total ways = 4*3*3 + 4*3*4 = 84

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

venky  (9 years ago)

0,1,2,3,4,5 divisible by 4
Ans:
_ _ _ _
last two digit should be divisible by 4 w/o repeat..
so first write possibles zeros and other number
Case:1
04,20,40
__ __ [__ __]
last two places 3 ways remaining two places 4*3ways
so totally 4*3*3=36;
Case:
except zero 12,24,32,52
__ __ [__ __]
last two digits in 4 ways and remaining 2 places are 4*3
totally 4*3*4=48
so Case1 and Case2
36+48=84 ways

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

amit kumar  (9 years ago)

the last two digits should be divisible by 4 in ordeer to complete divisibility of the number.
so, the possibilties are 04, 20,40,12,24,32,52
so total is (36+48) = 84

Is this solution Helpfull? Yes (2) | No   

ganesh babu  (9 years ago)

9*4+2*12=60
Last two digit possible 24,12,32,52,20,40

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

RAVISH KUMAR  (9 years ago)

ans 225
for thousand place we have 5 digits
for hundred place 5 digit
for last 2 places i.e ten and unit place
00
12
04
20
24
32
40
44
52
9 numbers which are divisible by 4
so total no of ways=5*5*9=225

Is this solution Helpfull? Yes (1) | No (4)  

Sameer Mohanty  (9 years ago)

The divisibility rule 0f 4 states that sum of last two digit has to be divisible by 4.
so the combinations we get is:
(4,0), (0,4), (1,3),(2,2),(3,1),(4,4),(6,2),(2,6),(5,3),(3,5),(7,1),(1,7),(8,8),(9,7),(7,9)

As we have to find the four digits number.. so the digits can range from 1000 to 9999. So each of the above combinations will come 100 times from 1000 to 9999. eg: 1040,1140,1240............9940

as here number of combinations are 15.. total numbers are 15*100= 1500

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ajit kumar sharma  (8 years ago)

with repetition
case 1 : last two digit containg 0
00,04,20,40(4 ways)
5*6*4=120
case 2: last two digit without 0
12,24,32,44,52(5 ways)
5*6*5=120
answer:120+150=270.
Without Repetition :
case 1 : last two digits containg 0
04,20,40(3 ways)
4*3*3=36
case 2: last two digits without 0
12,24,32,52(4 ways)
4*3*4=48
answer:36+48=84..

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

suresh reddy  (9 years ago)

how@binesh

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ASHISH MUKHERJEE  (9 years ago)

Ans must be ->75

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ASHISH MUKHERJEE  (9 years ago)

Sorry my previous Ans ..
it should be 63.

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AISHWARY KUMAR LAD  (9 years ago)

answer will be 63
4*3*2*1 when in unit place we put zero+
3*3*3*1 when in unit place we put two+
3*3*2*1when in unit place we put four=63

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vikash kumar anand  (9 years ago)

4 digit number can be formed as:-
-First digit should not be zero.
So, four digits number should be. 5*5*4*3=300
And number should be divided by 4.
So,300/4=75;75/4=18;18/4=4;4/4=1.
Hence, answer should be. 75+18+4+1=98

Is this solution Helpfull? Yes | No (1)  

Vijay Pareek  (9 years ago)

ans is 72 we can make 8 groups and each group will give u 9 numbers


like 20

5620
3420


3*3and lat 2 digit will be 20

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abhishek  (9 years ago)

36 explain

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anand vijay  (8 years ago)

without repetition answer should be 72 as i agree with siddharth prakash singh solution.

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aswani bonga  (8 years ago)

last 2 digits can be 04,12,20,24,32,40,44,52 that is 8 ways
100th palce in 6 ways
1000s place in 5 ways
8*6*5=240

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