Latest TCS Aptitude Question SOLUTION: Let P be the product of any three consecutive positive odd integers each of which is less than 146. Then the largest integer dividing all such P is 1 6 3 7 15

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Let P be the product of any three consecutive positive odd integers each of which is less than 146. Then the largest integer dividing all such P is

1
6
3
7
15

Asked In TCS deepak gupta (11 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

avinash trivedi  (11 years ago)

Anurag..what about 17*19*21,its not divisible by 15..
(17*19*21)%3 ==0 and divisible by 7..
(23*25*27)%3==0 but not divisible by 7
and so on...
so ans is 3...

Is this solution Helpfull? Yes (38) | No (5)  

Anurag  (11 years ago)

15 is the ans...
let the three nos. be a,a+2,a+4
1st we take lower case i.e a=1,b=3,c=5
So a*b*c=15 which is divisible by 15
Now we take the upper case a=141,b=143,c=145
So again a*b*c=2923635 which is divisible by 15

Is this solution Helpfull? Yes (19) | No (23)  

goldy  (11 years ago)

anurag what about 7,9,11=693,i;e not divisible by 15
thus correct answer is 3......

Is this solution Helpfull? Yes (5) | No   

juhi sharma  (11 years ago)

plss write the correct ans

Is this solution Helpfull? Yes (3) | No   

Mayank Dixit  (9 years ago)

Ans will be 3, because if we consider any possible consecutive odd numbers then one of them will be divisible by 3 thus making theri product divisble by 3.
eg {1,3,5} or {17,19,21} or {53,55,57} or {101,103,105} or {141,143,145}
3, 21, 57, 105, 141 all are divisble by 3 thus making their product divisble by 3

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