Latest Elitmus Aptitude Question SOLUTION: How many 4 digit can be formed by using 2,3,4,5,6,7 only once divisible by 25 remember only once* divisible by?? A)40 B)12 C)20 D)24

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(#M40056967) ELITMUS Numerical Ability question Permutation and combination Keep an EYE

How many 4 digit can be formed by using 2,3,4,5,6,7 only once divisible by 25
remember only once* divisible by??

A)40
B)12
C)20
D)24

Asked In Elitmus Shreya Banerjee (9 years ago)

Unsolved Read Solution (20)

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Read Solution (20) :

SMRITI  (9 years ago)

25 and 75 both are divisible by 25 once.
so for 25 as last two digits, no. of 4 digit number formed = 4*3*1*1 = 12
and for 75 as last two digits , no. of 4 digits number formed = 4*3*1*1 = 12
therefore , total number is 12+12 = 24

Is this solution Helpfull? Yes (24) | No (2)  

Pallavi Gupta  (9 years ago)

unit place will be only 5. tens place will be 2 or 7 only to be exactly divisible by 25 only once.
at hundred place=3 or 4 or 6. so there are total 3 ways and at thousand place there is only 2 ways.
so total possibilities = 2*3*2*1 = 12

Is this solution Helpfull? Yes (19) | No (17)  

Anand  (9 years ago)

units place= 5
tens place= 2 or 7(as tens place must have 2 or 7 to be exactly divisible by 25)
remaining places= 3 or 4 or 6 or one from 2/5
so total numbers formed= 3*2*2*2*1=24

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

SUMIT KUMAR  (9 years ago)

5+4+2+1=12

Is this solution Helpfull? Yes (8) | No (6)  

anil  (9 years ago)

20 all 24 once 24-4

Is this solution Helpfull? Yes (3) | No   

RUPA KUMARI  (9 years ago)

repeation is nt allowed hence... _ _ _ _
only divisivble if once is filled with 5
ten,s will b filled with 4 ways
similarly 100th posisition 3 ways n
1000th 2 ways _ _ _ _ 2*3*4=24

Is this solution Helpfull? Yes (3) | No   

RAVISH KUMAR  (9 years ago)

ans 12
last two digit must be 25 to be divisible by 25.
fix 2 and 5 and now we have 4 digit for rest of two places
for hundred place 4 digit
for thousand place 3 digit
so total number of numbers=3*4=12

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

Dipankar Dey  (9 years ago)

the numbers which are divisible by 2 times by 25 is 625*7 and 625*9..so the answer is 72-2=70...previous one was incorrect..

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

santu sahu  (8 years ago)

unit 2 place digit will be either 25 or 75 hence 2 ways
4th place can be puted (6-2) ways means 4 ways
3rd place can be puted as 3 ways
so 4*3*2=24

Is this solution Helpfull? Yes (1) | No   

kumar babu  (8 years ago)

The condition that the number is only once divisible by 25 means that it is a multiple of 25 but not 25^2=625. Since we are restricted to using the digits 2,3,4,5,6,7, the four-digit number will only be divisible by 25 if the last two digits are 25or 75. From these, we must remove those multiples of 625 between 2000 and 8000 in which the only digits are 2,3,4,5,6,7. Those multiples are:

2500,3125,3750,4375,5000,5625,6250,6875,7500

Assuming digits cannot be repeated, you correctly found the number of four-digit numbers that are divisible by 25. However, 4375 is also a multiple of 625 in which the only digits used are in the set {2,3,4,5,6,7} and no digits are repeated. Therefore, there are 24−1=23 numbers that satisfy the given condition.

Is this solution Helpfull? Yes (1) | No   

Dipankar Dey  (9 years ago)

the number of 4 digit numbers (inc 2,3,4,5,6,7) which are divisible by 25 is : 6 * 6* 2* 1
and the numbers which are two times divisible by 25 are also 1 time divisible by 625..so these numbers are only 625*5 , 625 *7, 625 *9.. because other numbers are either contains 0 or contain the digits which does not include in the question... so answer is (6*6*2*1-3)=69

Is this solution Helpfull? Yes | No (8)  

Abhishek singh  (9 years ago)

first 4*3and 25
second case 4*3 and 75
therfore 12+12=24 numbera

Is this solution Helpfull? Yes | No (1)  

ASHISH MUKHERJEE  (9 years ago)

Ans :-option b correct

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Bharath T S  (9 years ago)

3*4*2*1=24

Is this solution Helpfull? Yes | No   

vimal kumar joshi  (9 years ago)

**26 and **76
4c2=6
6+6=12
ans 12

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sheikh shahrukh  (9 years ago)

kindly mention if repetition is allowed or not....

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kaustubh salunke  (9 years ago)

24 is the ans

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Gazal  (8 years ago)

there are 4 positions _ _ _ _
unit digit will be only 5 so = only 1 way
tens place will be 2 0r 7 = 2 ways
so one digit at unit place and one digit at tens place
total remaining digits are 4
so remaining two positions can be filled in 4*3 ways
so answer is 4*3*2*1= 24

Is this solution Helpfull? Yes | No   

vikash rawat  (8 years ago)

at ones place 5 is fixed.
case 1. _ _25
in this case last two digit is fixed as 25.
so we have 3,4,6,7 options for remaining place
so no. of ways to fill first two place using 3,4,6, 7= 4*3=12

case2. _ _ 75
similarly in this case last two digit is fixed as 75.
so no. of ways to fill remaining two place is 4*3=12

final answer= case1 + case2 =12+12 =24

D is the answer

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sanjana  (8 years ago)

Last digit should be 5 so and tens digit can be either 2 or 7 so 2 ways and rest is 3*2*2=12

Is this solution Helpfull? Yes | No (1)  

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