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# If N = 7^2014 Then:A) The Last Digit is 7B) Remainder upon division by 24 is 1C) Remainder upon division by 25 is 1D) No. Of digits in N > 1700

9 months ago

2074 Points

N= 7^(2*1007)= (7^2)^(1007)= 49^1007
= (48+1)^1007
= 48k + 1 (using binomial theorem, k is an integer).
= 24(2k)+1
= 24q + 1
hence, the remainder is 1 when divided by 24.
for remainder by 25, write 49= 50 – 1= 25(2) – 1,and you ll find N= 25q – 1= 25(q – 1) + 24, hence remainder is 24 not 1.
for last digit, we simply need to find remainder on division by 10, write 49= 50 – 1= 10*5 – 1 and then proceed.
KINDLY APPROVE :))
9 months ago
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