Find the remainder when 1 2 5 0 0 0 1 9 7 6 8 3 is divided by 17500.
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By Fermat, we have 1 2 5 0 0 0 1 9 7 6 8 2 ≡ 1 ( m o d 7 ) , since the exponent is divisible by 6. Multiplying through with 1 2 5 0 0 0 = 5 0 × 2 5 0 0 , we find that 1 2 5 0 0 0 1 9 7 6 8 3 ≡ 1 2 5 0 0 0 ≡ 2 5 0 0 ( m o d 1 7 5 0 0 )
Normal remainder property I felt instead of using any theorems as such.
We can simplify the denominator with numerator value to find remainders in general.
1 2 5 0 0 0 1 9 7 6 8 3 / 1 7 5 0 0 = 5 0 1 9 7 6 8 3 / 7 = 1 .
125000/17500 = 50/7 which gives remainder as 1 and 1 power anything is always 1.
So, actual answer will be 1 * 2500 = 2500.
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Since the base of the given number is multiple of 50 , we set a simple congruence (for our ease) , 5 0 ≡ 1 m o d 7
Now raising the powers of both sides of congruence as follows :
( ( 5 0 7 ) 7 ) 7 = 5 0 3 4 3 ≡ 1 m o d 7
And at this point , We take Ramanujan's Number !
i.e , 5 0 3 4 3 . 1 7 2 9 ≡ 1 m o d 7
Now multiplying the whole congruence , by 5 0 2 = 2 5 0 0
We get , 5 0 5 9 3 0 4 9 ≡ 2 5 0 0 m o d 1 7 5 0 0 and hence required remainder is
2 5 0 0