AbhiroopGhosh wrote:
BunuelVeritasKarishmaThanks for your insights on the post.
Wanted to check if we can split the terms and find the individual remainders via the multiplicative theorem. Can you please let me know if that would be a correct approach.
For example -
We are asked to find the Remainder of (\(\frac{ 2^ {99} }{ 99}\))
(\(\frac{ 2^ {99} }{ 99}\)) = (\(\frac{ 2^ {99} }{33 }\)) * (\(\frac{ 1 }{ 3}\))
Therefore,
Remainder (\(\frac{ 2^ {99} }{ 99}\)) = Remainder (\(\frac{ 2^ {99} }{33 }\)) * Remainder (\(\frac{ 1 }{ 3}\))
Step 01 - Remainder (\(\frac{ 2^ {99} }{33 }\))
= Remainder [ (\(\frac{ 2^ {95} }{33 }\)) * (\(\frac{ 2^ {4} }{33 }\)) ]
= Remainder [ (\(\frac{ (2^ {5}) ^ {19} }{33 }\)) * (\(\frac{ 2^ {4} }{33 }\)) ]
= (\(\frac{ (-1) ^ {19} }{33 }\)) * 16
= (\(\frac{ (-1) ^ {19} }{33 }\)) * 16
= -16
Now we cannot have -16, therefore the remainder = 33 - 16 = 17
Remainder (\(\frac{ 2^ {99} }{ 33}\)) = 17
Step 02 - Remainder (\(\frac{ 1 }{3 }\))
= 1
Step 03 - Multiply both the remainders
= 17 * 1 = 17
This is not correct. Think about it: What if you were to divide 34 by 99? What would be the remainder?
Can you say it will be the same as remainder of 34/33 * 1/3 i.e. remainder will be 1? No. Remainder will be 34
The method used by
chetan2u uses some ingenuity. He observed that the remainders in the options were all less then 33.
When you divide n by 99 and if the remainder is less than 33, the remainder will be the same when you divided n by 33 because 33 is a factor of 99.
If this is not clear, think in terms of the grouping method. When you divide n into groups of 99 balls each, say r balls are leftover so remainder is r.
Now if r is less than 33, then even if you divided n into groups of 33 balls, the remainder will be the same i.e. r. All groups of 99 balls will just get split into 3 groups of 33 balls each.
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