But it is 1000 times!

Find the number of positive integers n n such 20 n + 2 20n+2 can divide 2003 n + 2002 2003n + 2002 .


The answer is 0.

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3 solutions

We want to find the number of positive integers n n such that 2003 n + 2002 20 n + 2 = m Z \dfrac{2003n+2002}{20n+2}=m\in \mathbb{Z} Simple polynomial division gives: m = 2003 n + 2002 20 n + 2 = 2003 20 + 1801.7 20 n + 2 20 m = 2003 + 18017 10 n + 1 \begin{aligned} m=\dfrac{2003n+2002}{20n+2}&=\dfrac{2003}{20}+\dfrac{1801.7}{20n+2}\\ \implies 20m &=2003+\dfrac{18017}{10n+1} \end{aligned} On factorizing, 18017 = 43 × 419 18017=43\times 419 .Therefore, it's only factors are 1 , 43 , 419 , 18017 1,43,419,18017 .None of these is of the form 10 n + 1 10n+1 .Hence there is no positive integer n n which will make 2003 n + 2002 20 n + 2 \dfrac{2003n+2002}{20n+2} and consequentially m m an integer.

Nice but I feel like I cheated a bit by using Python to find 18017 = 43 × 419 18017 = 43 \times 419 .

Arulx Z - 5 years, 2 months ago

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No problem :P

Abdur Rehman Zahid - 5 years, 2 months ago

By the way, isn't it enough to show that there is no integer n n such that 20 n + 2 20n+2 divides 1801.7 1801.7 because 1801.7 is not a integer while 20 n + 2 20n+2 is?

Arulx Z - 5 years, 2 months ago

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Yes that's a shortcut

Abdur Rehman Zahid - 5 years, 2 months ago
Syed Hamza Khalid
Sep 16, 2018

Let Q = 2003 n + 20002 20 n + 2 = 100 ( 20 n + 2 ) + 1802 20 n + 2 = 100 + 1802 20 n + 2 Q = \dfrac {2003n + 20002}{20n + 2} = \dfrac {100(20n+2)+1802}{20n+2} = 100 + \dfrac {1802}{20n+2} . For Q Q to be a positive integer 1802 20 n + 2 = 2 × 17 × 53 20 n + 2 = 2 × 17 × 53 2 × ( 10 n + 1 ) = 17 × 53 10 n + 1 \dfrac {1802}{20n+2} = \dfrac {2\times 17 \times 53}{\color{#3D99F6}20n + 2} = \dfrac {2\times 17 \times 53}{\color{#3D99F6} 2 \times (10n +1)} = \dfrac {17 \times 53}{\color{#3D99F6} 10n +1} must be a positive integer.

It is simple to notice that there is no value for n which would satisfy the conditions, hence the answer is 0 \boxed{0} .


Disclaimer: This question seems similar to my question

Puneet Pinku
Mar 30, 2016

A simple question which can be solved by polynomial division and knowing the fact that if number is divided by its factor the remainder is zero.The given dividend is 2003n +2002 and divisor is 20n+2. On dividing the dividend by divisor the remainder obtained is 3n+1802. Since the question asks us to find such a "n" such that the divisor is factor of the dividend, which implies the remainder = 0, 3n + 1802 =0, => n= -600.66667 . But the question asks for the number of positive integers and our answer is negative as well as not an integer. Hence, clearly the answer is zero.....

-266 doesn't satisfy. I think the only integral value is -42.

Kushagra Sahni - 5 years, 2 months ago

How did you get the remainder as (3n+798) ???

Chirayu Bhardwaj - 5 years, 2 months ago

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Good point out. Somehow by faulty subtraction I got 798 as answer. Thankyou very much for pointing it out. But it did not affect our answer n is still negative while the question asks for positive integers... I have edited the numbers in my answer.

Puneet Pinku - 5 years, 2 months ago

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