Ramanujan's Number at the top!

Find the remainder when 12500 0 197683 125000^{197683} is divided by 17500.


The answer is 2500.

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3 solutions

Since the base of the given number is multiple of 50 , we set a simple congruence (for our ease) , 50 1 m o d 7 50\equiv 1 \mod 7

Now raising the powers of both sides of congruence as follows :

( ( 5 0 7 ) 7 ) 7 = 5 0 343 1 m o d 7 ((50^{7})^{7})^{7}=50^{343}\equiv 1 \mod 7

And at this point , We take Ramanujan's Number !

i.e , 5 0 343.1729 1 m o d 7 50^{343.1729}\equiv 1 \mod 7

Now multiplying the whole congruence , by 5 0 2 = 2500 50^{2}=2500

We get , 5 0 593049 2500 m o d 17500 50^{593049}\equiv 2500\mod 17500 and hence required remainder is

2 5 0 0 {\color{#20A900} 2{\color{#20A900} 5{\color{#20A900} 0{\color{#20A900} 0}}}}

Otto Bretscher
Feb 20, 2016

By Fermat, we have 12500 0 197682 1 ( m o d 7 ) 125000^{197682}\equiv 1 \pmod{7} , since the exponent is divisible by 6. Multiplying through with 125000 = 50 × 2500 125000=50\times 2500 , we find that 12500 0 197683 125000 2500 ( m o d 17500 ) 125000^{197683}\equiv 125000\equiv \boxed{2500} \pmod{17500}

Ashwin K
Feb 20, 2016

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.

12500 0 197683 / 17500 = 5 0 197683 / 7 = 1 125000^{197683}/17500 = 50^{197683}/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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