P is a prime number greater than 3. The remainder of P² dividee by 24 must be.....?
Please solve it...and give a optimum solution.
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digit sum of any perfect square is 14970&divisibility rule of 3 is sum of digits should be divisible by 3 when a digit sum ofsquare of any prime number divided by 3 you get reminder 1
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prime numbers greater than 3 are always of the form of 6k+1 or 6k-1. put this in place of p. u will get 1 as remainder always
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answer will be 1. let's say a prime no 5 (3) 5^2=25 25/24=1 (remainder) if we take 7(3) 7^2=49 49/24=1( 49- 2*24= 49- 48=1 remainder)
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