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27 Jul 2023, 20:14
Kudos
Bookmarks
Bunuel
2^99 ÷ 99
= (2^9)^11 ÷ 99
= 512^11 ÷ 99
= 17^11 ÷ 99
= [(17^2)^5 * 17] ÷ 99
= [289^5 * 17] ÷ 99
289 ÷ 99 leaves remainder -8
Thus, we have (each stage representing equivalent expressions in terms of the remainders):
[(-8)^5 * 17] ÷ 99
= - [2^15 * 17] ÷ 99
= - [2^9 * 2^6 * 17] ÷ 99
= - [512 * 64 * 17] ÷ 99
= - [17 * 64 * 17] ÷ 99
= - [289 * 64] ÷ 99
= - [-8 * 64] ÷ 99
= 512 ÷ 99
= 17
Answer A
Alternatively:
99 is the LCM of 9 and 11
Let us find the remainder for 2^99 ÷ 11:
2^10 ÷ 11 leaves remainder 1
=> 2^99 ÷ 11
= [(2^10)^9 * 2^9] ÷ 11
= (1 * 512) ÷ 11
= 6 remainder
Let us find the remainder for 2^99 ÷ 9:
2^9 ÷ 9 leaves remainder -1
=> 2^99 ÷ 9
= [(2^9)^11] ÷ 9
= (-1)^11 ÷ 9
= -1 + 9
= 8 remainder
Thus, we have a number that leaves remainder 6 when divided by 11 and remainder 8 when divided by 9
The least such number, and hence, our required remainder, is 17
Answer A
21 Feb 2025, 10:06
Kudos
Bookmarks
According to powers 2^99 would have the last digit as 8 and any power of 99 would be either 9 or 1 so 8-9 would also give me 7 from subtracting 17-9 and 8-1 would also give me 7. IMO A
