Latest Others Aptitude Question SOLUTION: What is the largest positive integer n for which n3 + 100 is divisible by n+10?

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What is the largest positive integer n for which n3 + 100 is divisible by n+10?

Asked In Others Prerna (12 years ago)

Solved SDeep Bhattacharya Read Solution (4)

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SDeep Bhattacharya  (12 years ago)

(n^3 + 100)=[(n^3 + 1000)-900]=[(n^3 + 10^3)-900]....[eqn(1)]

Now,clearly (n+10) is a factor of (n^3 + 10^3),i.e,(n+10) clearly divides

(n^3 + 10^3) for any +ve integer n..

So,from [eqn(1)],we can conclude that (n+10) must divide 900,and then only

(n^3 + 100) will be divisible by (n+10)..

Now at this point,it is very easy to calculate the largest value of n for which

(n+10) will divide 900.

The largest value of n, (n+10)=900; =>n=890.

So,largest positive integer n for which (n^3 + 100) is divisible by (n+10) is

=890..

Thanks to all..:-):-)

Is this solution Helpfull? Yes (17) | No (1)  

PARAG KAMRA  (12 years ago)

By the division algorithm, n3+100 = (n+10)(n^2-10n+100)-900,
so
(n3 + 100)/
(n + 10)
= {(n+10)(n^2 - 10n + 100) -900}/(n + 10)
= (n^2-10n+100)-(900/n+10)
:
If n3 + 100 is to be divisible by n + 10, then 900=(n + 10) must be an integer.
Thus, the largest possible n for which this is true is n = 890.

Is this solution Helpfull? Yes (5) | No (4)  

Prerna  (12 years ago)

What is the largest positive integer n for which (n^3 + 100) is divisible by (n+10) ?

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

laxmi  (10 years ago)

Ans: n = 0
The digit sum of 4444 is when remainder obtained 4444 divided by 9
4444 = (45−1)44
Each term is a multiple of 9 but the last term which is (−1)44 = 1
So the digit sum of 4444 is 1.

Now the divisibility rule for 3, 9, 27... is the sum of the digits should be divisible by 3, 9, 27 respectively. In each case the digit sum is either multiple of 3 or 9.
So for any value of n > 1, the given expression is not divisible by 3n

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