prove that for any natural number n A=2903^n-803^n-464^n+261^n is divisible by 1897?

Answer 1

I'm going to prove this by repeated induction. Well that's what I call it.

Let f(n) = 2903^n-803^n-464^n+261^n
You can verify that f(1) = 1897.
Assume that f(k) is div by 1897
So let's investigate f(k+1)

2903*f(k) = 2903^k+1 -803^k+1 - 2100*(803)^k -464^k+1 - 2439*(464)^k +261^k+1 + 2642*(261)^k
2903*f(k) = f(k+1) + g(k)
where g(k) = - 2100*(803)^k - 2439*(464)^k + 2642*(261)^k

So it follows that if g(k) is divisible by 1897, the f(k+1) is div by 1897. Let's prove g(k) by induction.
g(1) = -1122*1897
Following the same pattern as above
803*g(k) = g(k+1) + h(k)
where h(k) = -826821*(464)^k + 1431964*(261)^k

Again if h(k) is div by 1897, the g(k+1) is as well.
h(1) = -5220 * 1897
464*h(k) = h(k+1) + 203*1431964*(261)^k
= h(k+1) + 1897 * 153236 * (261)^k

So it is evident that h(k+1) is div by 1897
thus h(n) is div by 1897 for all n,
hence g(n) is div by 1897 for all n, and
hence f(k+1) is div by 1897,
hence f(n) is div by 1897 for all n.

Proved by induction ... eventually.

Answer 2

Factor 1897 = (7)(271) Reduce:

2903 = 5mod 7, 803 = 5mod7, 464 = 2mod7, 261 = 2mod7

Therefore A = 5^n - 5^n - 2^n + 2^n =0 mod 7.

Reduce: 2903 = 193mod271, 803 = 261mod271,

464 = 193mod271, 261=261mod271. As before,

A = 193^n - 261^n - 193^n + 261^n = 0 mod 271.

Since the primes 7, 271 divide A for all n, the product

(7)(271) = 1897 divides A for all n.

Answer 3

Notice that
2903^n-803^n-464^n+261^n is congruent to 193^n-261^n-193^n+261^n=0 modulo 271
and
2903^n-803^n-464^n+261^n is congruent to 5^n-5^n-2^n+2^n=0 modulo 7,
hence 271 divides A and 7 divides A. Since 271 and 7 are prime, this means 271*7=1897 divides A.

Answer 4

use mathematical induction
p(1)=2903-803-464+261=1897
p(2)=8427409-644809-215296+68121=7635425=4025*1897
so let p(n-1)=1897*k or a multiple of 1897 where k is a natural number & n>=1
so 2903^(n-1)-803^(n-1)-464^(n-1)+261^(n-1)=1897*k
so p(n)=2903^n-803^n-464^n+261^n must be a multiple of 1897
p(n-1)*2903=1897k*2903=2903^n-2903*803^(n-1)-2903*464^(n-1)+2903*261^(n-1)
p(n)=1897k*2903+2903*803^(n-1)+2903*464^(n-1)-2903*261^(n-1)-803^n-464^n+261^n
=1897k*2903+(2903-1)803^(n-1)+(2903-1)464^(n-1)-(2903-1)261^(n-1)
=1897k*2903+2902[803^(n-1)+464^(n-1)-261^(n-1)]
=1897k*2903+2902[2903^(n-1)-1897k)]
=1897k(2903-2902)+2902*2903^(n-1)
=1897k+2902*2903^(n-1)

Answer 5

prove it by mathematical induction...first piuut n=1...u get1897...which is divisble...hen for any value of n=k....get an equation...
then put n=k+1....
and use the equation u got in last step to prove its divisibilliy...its easy enough....
ur homework:do the last 2 steps...lol...good question man..wil help u to introspect ur concepts....
hope it helped..