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If n be a odd number greater than 1. than n(n*n-1) will be divisible by................ a)48 b) 24 c)6 d) none of these
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- Odd number just greater than 1 is 3
n(n*n-1)
3(3*3-1)
3(9-1)
3(8)
24
So it is dvisible by both 6 and 24 - 12 years agoHelpfull: Yes(18) No(4)
- option (d ) is correct
lets check it
on 1 n(n*n-1)= 1(1*0)=0
for 3 =3 (3*2)=18 divisble by 6
for 5 =5 (5 * 4)=100 not divisble by 6
now by option if a number is divisble by 48 means it would be divisble by 24,6 as well
if we take 24 then option c will also be true
thats why we check only for 6 and found not always divisble number so option (d)
is correct - 12 years agoHelpfull: Yes(6) No(35)
- The answer is 6 and 24
- 12 years agoHelpfull: Yes(5) No(2)
- taking n=7 48,24 and 6 all are answer....i think its a dummy question
- 12 years agoHelpfull: Yes(3) No(0)
- i think d option will be all of the above not none of these
- 12 years agoHelpfull: Yes(2) No(1)
- 6 and 24 both
as we can suppose minimum odd no. 3
and we substitute its value in equation
we get 24
- 11 years agoHelpfull: Yes(1) No(0)
- n(n*n-1)
= n(n^2-1)
=n(n+1)(n-1)
=(n-1)n(n+1)
now n is odd so (n-1)&(n+1) is even..we can get 2*2=4 from it..
and there are three consicutive nos so..one of them is multiple of 3 so we will get one 3 also
all we get two 2's one 3
so it have to be a multiple of 2*2*3=6 (ans)
- 11 years agoHelpfull: Yes(0) No(0)
- n(n*n-1)
= n(n^2-1)
=n(n+1)(n-1)
=(n-1)n(n+1)
now n is odd so (n-1)&(n+1) is even..we can get 2*2=4 from it..
and there are three consicutive nos so..one of them is multiple of 3 so we will get one 3 also
all we get two 2's one 3
so it have to be a multiple of 2*2*3=24 (ans) - 11 years agoHelpfull: Yes(0) No(0)
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