last two digit

Question

What would be last two digit of 25⁶³ x 63²⁵ .

Answer 1

Use mod(y,x) as y mod x

We only need to consider the last 2 digits of each of 25^63 and 63^25

25^n, n >= 1, has a final 2 digits of 25 for all n  (easily proven by induction, as 25 * 25 = 625 has last 2 digits 25)

Then, the last 2 digits of 25 * y = the last 2 digits of 25 * mod(y, 4) - the pattern is 25, 50, 75, 0.

Thus, we only need to find what mod(63^25, 4) equals.

mod(63, 4) = 3

All numbers equal to 3 mod 4 have the power pattern mod 4 of 3, 1, 3, 1, ... (3^2 = 9, and mod(9, 4) = 1)

Thus, mod(63^n, 4) = 3 if n is odd and 1 if n is even.

25 is odd.

Thus, mod(63^25, 4) = 3.

Thus, mod(25^63 * 63^25) = 75 mod 100

 

Answer 2

Take the modulus to be 100.

 

For 25, 25ⁿ ≡ 25 (mod 100) for any integer n.

 

To show this let n =1 then 25 ≡ 25 (mod 100)

Assume true for n = k, 25^k ≡ 25 (mod 100)

Then 25^k+1 = 25^k * 25 ≡ 25 * 25 = 625 ≡ 25 (mod 100)

This show the result by mathematical induction.

 

Also 63 = 7 * 9

 

Then 7⁵ = 16807 ≡ 7 (mod 100)

7²⁵ = (7⁵)⁵ ≡ 7⁵ ≡ 7 (mod 100)

 

Also 9⁵ =59049 ≡ 7² (mod 100)

9²⁵ = (9⁵)⁵ ≡ (7²)⁵ = (7⁵)² ≡7² = 49 (mod 100)

Putting all the pieces together,

 

25⁶³ * 63²⁵ = 25⁶³ * 7²⁵ * 9²⁵

                   ≡25 * 7 * 49 (mod 100)

                   ≡ 8575 (mod 100)

                   ≡75 (mod 100)

 

The last two digits of the product is 75