Question

Prove that for any natural number. n E=2903^(n)-803^(n)-464^(n)+261^(n) is divisible by 1897​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Let us first observe that 1897=271×7. Since 271 and 7 are prime to each other (actually they are both primes), therefore it is enough to show that A is divisible by 7 as well as by 271. Let us write A=(2903

n

−803

n

)−(464

n

−261

n

)

Since a

n

−b

n

is always divisible by a−b, therefore (2903

n

−803

n

)− is divisible by 2903−803=2100 and 2903

n

−803

n

is divisible by 2903−803=2100 and 464

n

−261

n

is divisible by 264−261, i.e., 203. Since 2100 and 203 are both divisible by 7. it follows that A is divisible by 7.

Again let us write

A=(2903

n

−464

n

)−(803

n

−261

n

)

Now 2903

n

−464

n

is divisible by 2904−464i.e.,2439 which is a multiple of 271,so that 2903

n

−464

n

is divisible by 271.

Also 803

n

−261

n

is divisible by 803−261 i.e., 542 which is multiple of 271, so that

803

n

−261

n

is a multiple of 271.

Since 2903

n

−464

n

and 803

n

−261

n

are both multiples of 271, therefore A is a multiple of 271. Since A is a multiple of both 271 and 7 it follows that A is a multiple of 271×7,i.e.,1897