Latest Elitmus Aptitude Question SOLUTION: how many four digit number can be formed using 2,3,5,6,7 and 8 which is divisible by 25.

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(#M40033532) ELITMUS Numerical Ability question 2nd march question Keep an EYE

how many four digit number can be formed using 2,3,5,6,7 and 8
which is divisible by 25.

Asked In Elitmus dhanraj kumar (10 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

dhanraj kumar  (10 years ago)

the number will be 25 and 75 in last digit
so for 25 as last digit= 4*3*1*1=12
for 75 as a last two digit= 4*3*1*1= 12
so, 12+12=24

ans=24

Is this solution Helpfull? Yes (34) | No   

pankaj kumar  (10 years ago)

the no in last degit shuld be (00, 25, 50, 75) which is divisible by 25
but 0 is not persent so last digit(25, 75)
then the selection of first digit is fix 5 =1
then the selection of second digit is (2,7)=2
then the selection of third and four digit is = 4*3
so total no. of probability= 4*3*2*1 = 24

Is this solution Helpfull? Yes (10) | No (1)  

deep chandra bhatt  (10 years ago)

ques ye tha how many four digit can be formed using 2,3,4,5,6,7 only once divisible by 25
a 12
b 20
c 24
d 40

Is this solution Helpfull? Yes (4) | No   

Mani kandan  (10 years ago)

Ans : 24

For any number to be divided by 25, it should be ended with {25,50,75,100)...
Here 0 is not in the list... so d possibilities are the numbers ended with (25 nd 75)

if v take the no. to end with 25, the remaining 2 digits can be any 2 among {3,4,6,7} . so the possibities are 4!/2= 24/2= 12

if v take the no. to end with 75, the remaining 2 digits can be any 2 among {2,3,4,6} . so the possibities are 4!/2= 24/2= 12

so 12+12=24

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deep chandra bhatt  (10 years ago)

24 ans nahi ho skta because 4375 divide 25 two times

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dhanraj kumar  (10 years ago)

sorry its using 2,3,4,5,6,7

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MANISH KUMAR  (10 years ago)

---5 the no will divisible by 25 if the last digit of the no must be 0 or 5. so first we fix the no 5 at fourth place of the no then we arrange the numbers at remaining three places out of six remaining number 6p3=!6/!6-3=60 ans

Is this solution Helpfull? Yes | No (3)  

mukul jain  (10 years ago)

the no which can be divided by 25 always have 5 at the end and ten digit should be either 2 or 7
and the remaining palaces will be fill by 3,6,8
so
3*2*2*1=12 ans

Is this solution Helpfull? Yes | No (3)  

ARPIT SRIVASTAVA  (10 years ago)

Its 24...try fixing last two digits by which can b only 25 n 75..

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srija  (10 years ago)

sry sryyy... if it is using 2,3,4,5,6,7 then
last two positions= 2 ways
remaining places = 6*6(repetition allowed)
so total no.of way= 2*6*6=72

Is this solution Helpfull? Yes | No (1)  

pawan  (9 years ago)

first luck for the divisibility of 25...
no will be. _ _ _ _ 5
_ _ _ 7 5
now three place can we fill in 4*3*2 way....ans is 24

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suvarna  (9 years ago)

two nos will be divisible by 25; 25,75.
--(25or75) then 4p2*2p1=24

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