Question

what is the value of n for which (201^n)- (152^n) is divisible by 2009?

Answer 1

Answer:

n=0

Step-by-step explanation:

Using the formula $x^{n} -k^{n} =(x-k)[x^{n-1}+(x^{n-2})k.....k^{n-1}$, here x=201 and k=152 and x-k is constant irrespective of n.

$201^{n}-152^{n} =49 (-)$ which is divisible by 2009,

therefore, 2009=49×41

Hence, (-) should be divisible by 41.

Now, 152=8×19 and 201=3×67, the common divisor of both the numbers is one, therefore, 201-152=(1)(3)(67)-(1)(8)(19)

                                      =(1){3×67-8×19}

Now, $201^{2}= (1)(3)(3)(67)(67)$ and $152^{2} =(1)(8)(8)(19)(19)$, Again, the common divisor of both these numbers remains one and further, for any power of 201 and 152, the common divisor will be 1.

Therefore, by the definition of prime numbers, the resultant number of $201^{n} -152^{n}$ will be prime number(41 is a prime number) and is not divisible by any other prime number, therefore, n=0 is the only solution.

Hence, for n=0, $201^{n} -152^{n}$  will be divisible by 2009.