Question

[{(6499)^6 (9841)^6 ] - 2 } is divisible by 19. Prove this

Answer 1

Answer:

[{(6499)^6 + (9841)^6 ] - 2 } is divisible by 19

Step-by-step explanation:

we will split the two digits 6499 and 9841 in terms of multiples of 19 and remainder.

6499 = ( 342 × 19 ) + 1  ......  (1)

9841 = ( 518 x 19) - 1     ....... (2)

Let

342×19 = A    &

518×19 = B

where A and B are divisible by 19.

Substituting in equation 1 and 2 we get,

6499 = A+1 ...... (3)

9841 = B-1  ...... (4)

Substituting 3 & 4 in main equation we get,

[{(6499)^6 + (9841)^6 ] - 2 } = [{(A+1)^6 + (B-1)^6 ] - 2 }  ..... (5)

Now, any polynomial in the form (p+q)^6 can be expanded as

(p+q)^6  = p^6 + 6(p^5)(q) + 15(p^4)(q^2) +  20(p^3)(q^3) +  15(p^2)(q^4) +  6(p^1)(q^5) +  (q^6)

Thus,

(A+1)^6 = A^6 + 6(A^5)(1) + 15(A^4)(1^2) +  20(A^3)(1^3) +  15(A^2)(1^4) +  6(A^1)(1^5) +  (1^6)

(B-1)^6 = B^6 + 6(B^5)(-1) + 15(B^4)(-1^2) +  20(B^3)(-1^3) +  15(B^2)(-1^4) +  6(B^1)(-1^5) +  (-1^6)

From the above two equation we observe that, out of 7 terms expanded, 1st 6 terms are multiple of A in 1st and B in 2nd expansion.

Thus we can re-write the same as

(R and S are integers)

(A+1)^6 = R×A +  (1^6) = R×A + 1 ...... (6)

(B-1)^6 = S×B +  (-1^6) = S×B + 1 ....... (7)

Substituting in equation (5), we get

[{(6499)^6 + (9841)^6 ] - 2

= [{(A+1)^6 + (B-1)^6 ] - 2 }

= [{R×A + 1 + S×B + 1] - 2 }  

= R×A +1 + S×B + 1 - 2

= R×A + S×B

∴ we can simplify

[{(6499)^6 + (9841)^6 ] - 2  =  R×A + S×B

as A & B are divisible by 19

the whole number after calculation will be divisible by 19.

Thus Proved.