Latest IBM Aptitude Question SOLUTION: Put one digit before 15 and one digit after 15 and find out how many such numbers are divisible by 15

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Put one digit before 15 and one digit after 15 and find out how many such numbers are divisible by 15

Asked In IBM (10 years ago)

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Shubhangi  (10 years ago)

for a no to be divisible by 15 it should be divisible by both 3 and 5 bcoz dese are its prime factors.
A no could be divisible by 5 if it has 5 or 0 at the unit place.
So we can put only 5 or 0 at the unit place i.e. after 15
1)If we have put 5 then for a no to be divisible by 3 the sum of digits should be divisible by 3
so possible cases are
1+1+5+5=12,so we can put 1,similarly at thousand place we can put,4,7
So total possible cases when 5 is at unit place are 3.
2)When 0 is at unit place
Then we can put 3,6,9 at thousand place to make it divisible by 15.
Hence total cases and total nos divisible by 15 are 6
1155,4155,7155,3150,6150,9150.

Is this solution Helpfull? Yes (26) | No (2)  

lalita kesarwani  (10 years ago)

ans. 7
case 1: unit digit is 5 then we have 3 cases with 4,7,1 at thousands place
case2: unit digit is 0 then we have 4 cases with 3,6,9,0 at thousands place
hence total of 7 cases

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

Vishakha Butta  (10 years ago)

answer is 7.
numbers are 1155, 4155, 5155, 3150, 6150, 7150, 9150

Is this solution Helpfull? Yes (2) | No (8)  

kumar gaurav  (10 years ago)

Let us assume the number is x15y
This is divisible by 3 and 5.
As the number is divisible by 5 y should be either 0 or 5
As the number is divisible by 3 the sum of digits should be divisible by 3
x+1+5+y= x+y+6 is divisible by 3
Let us say x+y+6=3n
x+y=3n-6
=3(n-2)-----------------(1)
Case 1: y=0
Substituting in (1) we get x= 3(n-2)
n=3 x=3
n=4 x=6
n=5 x=9
n=6 x=12 not possible as x is supposed to have only one digit
The only possible one digit values of x satisfying this equation are 3, 6 and 9
The possible nos are 3150,5150,9150

Case 2:y=5
substituting in (1) we get x=3n-11
when n=4,x=1;
n=5,x=4;
n=6,x=7
n=7,x=10 not possible as x is supposed to have only one digit.
So the possible values of nos are 1155, 4155, 7155
So we get totally 6 possible nos 1155, 4155, 7155, 3150, 5150, 9150

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

Ravi Ranjan Raj  (10 years ago)

@shubhangi why can't use the number 0150 b'cos its not given the no. must be 4 digit

Is this solution Helpfull? Yes (1) | No   

saurabh raj  (10 years ago)

0150
3150
6150
9150
1155
4155
7155

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

shikha  (10 years ago)

5 as 315, 615, 915, 155, 150

Is this solution Helpfull? Yes | No (7)  

Neha Saxena  (10 years ago)

No to be divisible by 15-should be divisible by both 3 and 5.
A no could be divisible by 5 if it has 5 or 0 at the unit place.
So we can put only 5 or 0 at the unit place
1)If we have put 5 then for a no to be divisible by 3 the sum of digits should be divisible by 3
so possible cases are
1+1+5+5=12,so we can put 1,similarly at thousand place we can put,4,7
So total possible cases when 5 is at unit place are 3.
2)When 0 is at unit place
Then we can put 0,3,6,9 at thousand place to make it divisible by 15.
Hence total cases and total nos divisible by 15 are 7
1155,4155,7155,0150,3150,6150,9150.

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