Latest Elitmus Aptitude Question SOLUTION: How many three digit number formed by using 2,3,4,5,6,7 once such that number is divisible by 15. a) 8 b) 16 c)12 d)20

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(#M40157148) ELITMUS Numerical Ability question Keep an EYE

How many three digit number formed by using 2,3,4,5,6,7 once such that number is divisible by 15.
a) 8
b) 16
c)12
d)20

Asked In Elitmus ranjeet singh (7 years ago)

Unsolved Read Solution (27)

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Read Solution (27) :

Abhijeet Jain  (7 years ago)

If a no. is divisible by 15, so it is also divisible by 3 and 5
for a no. to be divisible by 5 the unit digit must be 0 & 5 but 0 is not given so unit digit is 5.
now use the divisibility test of 3 hence we get 8 different no.s which are
345 435
375 735
465 645
675 765

Is this solution Helpfull? Yes (36) | No (8)  

Payel Kayal  (7 years ago)

The rule of a number divisible by 15 is the number is end with 5 and the sum of digits is divisible by 3 also. Here 5 is fixed at the last position and remaining 5 digits can fill 10th and 100th place in 8 different ways such that
(3,4)
(3,7)
(4,3)
(4,6)
(6,4)
(6,7)
(7,3)
(7,6)
So the ans is 8.

Is this solution Helpfull? Yes (14) | No (2)  

Archit Agarwal  (7 years ago)

4*2*1=8
at one's place only 5 will come
and at ten's place 4 and 7 can be placed,
and at 100th place rest of the 4 digits can come...
so the answer is 8

Is this solution Helpfull? Yes (11) | No (10)  

sanchita dasgupta  (7 years ago)

unit digit must be 5 i.e _ _ 5 (only one number in unit place)
Now, we have 5 more number that can be in 10th place that are 2,3,4,6,7 (5 will not because 5 has already used )
so we have _ (any one from 2,3,4,6,7) 5
now we have only 4 numbers to be used on 100th place
so we 4*5*1=20

Is this solution Helpfull? Yes (7) | No (15)  

Sibabrat Swain  (7 years ago)

1- no divisible by 15 should be divisiblled by both 5 and 3 .
2-divisibility conditions for 5 is the unite place should be 5.
3-now we have to check how many numbers are there which will be divisible by 3 with unite place 5 of that no .
4-for check the no is divisible by 3 , the summation of digits should be divisible by 3 .
5- manually you can check
these option would be (4 3 5) (3 4 5)(5 2 5 )(2 5 2)(7 3 5)(3 7 5)(6 4 6)(4 6 4)(7 6 5)(6 7 5)(2 2 5)(555)
so there wil be 12 options .

Is this solution Helpfull? Yes (4) | No (20)  

Sibabrat Swain  (7 years ago)

THEN IT IS 8 BROTHER .#abhijit jain

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

Manojkumar  (7 years ago)

235 345 465 675 725
245 365 475 625 735
265 375 425 635 745
275 325 435 645 765 ans :20

Is this solution Helpfull? Yes (4) | No (52)  

mallikarjuna  (7 years ago)

8 is the correct answer. we cant use a digit more than once. so 8 will be the only valid answer here.

Is this solution Helpfull? Yes (3) | No   

Abhijeet Jain  (7 years ago)

sibabrat , in the question it is clearly mentioned that repetition of digits is not allowed.

Is this solution Helpfull? Yes (2) | No   

monika sinha  (7 years ago)

There are 6 number take it unit ,tens and hindered . We know that to divide by 15 unit should be 5 and it is 1way and remaining 5 will in two ways 5c2 than 20 is ans

Is this solution Helpfull? Yes (2) | No (3)  

vikram singh  (7 years ago)

if no. is divisible by 15 at once than last two digit must be
(15,30,45,60,75,90.......135)
but here only two choice available either 45 or 75.
so first case
when last two digits 45 then
no.of three digits =6
and when last two digits =75
then no. of three digits=6
add both=6+6
=12
ans=12

Is this solution Helpfull? Yes (2) | No (5)  

Himanshu Kumar  (7 years ago)

8
the no. formed which should be divisible by 15 means it must be divisible by 3 and 5
than unit place should be 5 (for divisibility of 5)
for divisibility of 3
345
435
375
735
465
645
675
765

Is this solution Helpfull? Yes (2) | No   

Sibabrat Swain  (7 years ago)

MONIKA divisibility criteria of 15 is not only unit place should be 5. Unite place is 5 then it is divisible by 5 only .

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PRATIBHA  (7 years ago)

ans: 8 because there is no repition , for divisible of unit place should be 5 and for divisible of 3 sum of all digit should b divisible by 3

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sushil khati  (7 years ago)

opt. c 12
divisible no. are
135,315,165,615,
345,435,375,735,
465,645,675,765

Is this solution Helpfull? Yes | No (8)  

s hotta  (7 years ago)

1- no divisible by 15 should be divisiblled by both 5 and 3 .
2-divisibility conditions for 5 is the unite place should be 5.
3-now we have to check how many numbers are there which will be divisible by 3 with unite place 5 of that no .
4-for check the no is divisible by 3 , the summation of digits should be divisible by 3 .
5- manually you can check
these option would be (4 3 5) (3 4 5)(5 2 5 )(2 5 2)(7 3 5)(3 7 5)(6 4 6)(4 6 4)(7 6 5)(6 7 5)(2 2 5)(555)
so there wil be 12 options .

Is this solution Helpfull? Yes | No (5)  

shubham pal  (7 years ago)

3p5=120
120/15=8

Is this solution Helpfull? Yes | No (1)  

anjali  (7 years ago)

which one is correct ans ? 8 or 20 ?

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shubham pal  (7 years ago)

sorry for the last answer i.e 5! i.e 120 /15 =8

Is this solution Helpfull? Yes | No   

Anshul Nagi  (7 years ago)

5 gets fixed in unit place now we require a sum of 4 ,7,10,13 using other digits to make it divisible by 3 (4+5=9 ,7+5=12.....) which satisfy divisibility by 3.
Now possible ways for
4 =(2,2)
7=(5,2) (2,5) ,(4,3) ,(3,4)
10=(5,5) ,(6,4) ,(4,6) , (7,3) ,(3,7)
13=(7,6) ,(6,7)
Total =12

Is this solution Helpfull? Yes | No (1)  

subhajit chakraborty  (7 years ago)

in order to make the number divisible by 15..the last digit should be 5 so that it is divisible by 5 ..and the summation of the numbers is (2+3+4+6+7+5=27)i.e divisible by 3 hence the last digit is kept fixed to 5 n the 1st 2 digits would be made with 2,3,4,6,7..so ans is5C2=20

Is this solution Helpfull? Yes | No   

pritam sinha  (7 years ago)

b) 8 is right answer
three digit number divisible by 15 so number must be divisible by 3 and 5. so 5 is fixed on unit digit . and we have to consider tenth and thousand place respectively so that number could be divisible by 3.
so pair is (3,4), (4,3), (3,7), (7,3), (4,6), (6,4), (6,7), (7,6).

Is this solution Helpfull? Yes | No   

MONDURI MAHESH  (7 years ago)

a number is divisible by 15 if last digit is either 5 or 0...so last digit can be filled in one way ie 5.other 2 digits can be formed from 5 digits in 5p2 ways.ie 20.

Is this solution Helpfull? Yes | No   

Sujata  (7 years ago)

Answer is 8,
If number is divisible by 15 then it should be divisible by both 3 and 5,
So, for a number to be divisible by 5, it's last digit should be either 5 or 0. Since 0 is not given, unit digit is 5,
now for a number to be divisible by 3, the sum of digits of number should be divisible by 3.
So, we can avoid numbers that include 2, so there are only 12 different combinations. Out of which we can eliminate 4 combinations (3,6,5), (4,7,5), (6,3,5), (7,4,5) as their digit summation is not multiple of 3. So, there are only 8 different combinations.

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sandeep  (7 years ago)

Any number which is divisible by 15 .
The number must be divisible by both 5 and 3
First we written numbers which are divisible by 5 ( because every number which is divisible by 5 , last must be zero or 5 )
5*4*1=20(total 20 numbers - - 5)
Now check in 20 numbers which is divisible by 3
Rule of sum of all digit of a number is n.
n%3=0
.......
........
check all numbers and after checking answer is ::8

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Zaki Anwer  (7 years ago)

If a number is divisible by 15 it must be divisible by both 3 and 5; bcz 3x5=15;
So, a 3 digit number divisible by 5 must have 5 or 0 at unit place : __ __ 5 (here 0 is unavailable)
a 3 digit number divisible by 3 must have sum of digits, divisible by 3.
The only options are : 3 4 , 3 7, 4 6, 6 7 ; i,e, total 4 options
all have two combinations
So 4x2 = 8
Ans : a) 8

Is this solution Helpfull? Yes | No   

jyoti shankar  (6 years ago)

In a three digit number, the last digit is always 5 and 0 but 0 is not given so ignore 0. Now the remaining 100ths place will be formed by 5 ways and 10ths place will be formed by 4 ways.
Finally, the result is 5*4*1 i.e., 20 ANSWER.

Is this solution Helpfull? Yes | No (1)  

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