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I wouldn’t say there is an easier way as such and may be you would get some more friendly terms than 99.
Having said that, you can ease it logically by seeing that all choices are 17 or less, so when you divide by 33 too, the remainder should be same.
Now \(2^{99}=(2^5)^{19}*2^4=32^{19}*2^4\) Now 32 will leave -1 as remainder So remainder = \((-1)^{19}*16=-1*16=-16\), or \(33-16=17\)
OR
What will be the remainder when 2^99 is divided by 9, a factor of 99 \(2^{99}=(2^3)^{33}=8^{33}\) 8 will leave a remainder 1, so \(R=(-1)^{33}=-1\) or \(9-1=8\).. Since we are looking for remainder with 99, remainder can be anything in the form \(9k+8\) So answer will be one of 8,17,23... Only 17 is there so our answer must be 17
A point on one of the solution that since 99=3*3*11, so remainder =3+3+11 is WRONG. As there are kudos given to solution, don’t take it as a correct method. The solution doesn’t even talk of 2^99, so anything 2^3 or 2^13 or 5^1 and so on will give 17 as answer _________________
Re: What is the remainder when 2^99 is divided by 99?
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02 Nov 2019, 12:47
5
2
[quote="Bunuel"]What is the remainder when \(2^{99}\) is divided by 99?
A. 17 B. 15 C. 13 D. 11 E. 9
since 99 can be written as 3*3*11, and none of the three prime factors can be cancelled with the numerator. we simply add the three primes. (3+3+11 = 17) answer - A) 17
What is the remainder when 2^99 is divided by 99?
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03 Nov 2019, 02:32
2
1
8=1*9-1
\(2^3=-1 mod 9\) \((2^3)^{33}=(-1)^{33}\) mod 9
\(2^{99}\)= -1 mod 9= 8 mod 9
\(2^{99}=9m+8\).......(1)
32=3*11-1 \(2^5\)=-1 mod 11 \((2^5)^{19}\)=-1 mod 11 \(2^{95}*2^4\)=-1*5 mod 11 \(2^{99}\)= -5 mod 11= (11-5) mod 11 \(2^{99}\)= 6 mod 11
\(2^{99}=11n+6\).......(2)
So basically our question stem is "The remainder when N is divided by 9 is 8, and when divided by 11 is 6. What is the remainder when N is divided by 99"?
from 1 and 2
\(2^{99}= lcm(9,11)x+17\) \(2^{99}= 99x+17\)
Bunuel wrote:
What is the remainder when \(2^{99}\) is divided by 99?
Hence removing multiples of 99, we are left with - \(\frac{(17)^{11}}{99}\) \(\frac{(17)^{10} * 17}{99}\) \(\frac{(17^{2})^{5} * 17}{99}\) \(\frac{(289)^{5} * 17}{99}\) \(\frac{(297 - 8)^{5} * 17}{99}\) \(\frac{(99*3 - 8)^{5} * 17}{99}\)
Hence removing multiples of 99, we are left with - \(\frac{(-8)^{3} * (-8)^{2} * 17}{99}\) \(\frac{(-512) * (-8)^{2} * 17}{99}\) \(\frac{(-495 - 17) * (-8)^{2} * 17}{99}\) \(\frac{(-(99*5) - 17) * (-8)^{2} * 17}{99}\)
Hence removing multiples of 99, we are left with - \(\frac{(-17) * (-8)^{2} * 17}{99}\) \(\frac{(-289) * (-8)^{2}}{99}\) \(\frac{(-297+8) * (-8)^{2}}{99}\) \(\frac{(-(99*3)+8) * (-8)^{2}}{99}\)
Hence removing multiples of 99, we are left with - \(\frac{8 * (-8)^{2}}{99}\) \(\frac{8 * 64}{99}\) \(\frac{512}{99}\) \(\frac{(495 + 17)}{99}\) \(\frac{(99*5 + 17)}{99}\)
Hence removing multiples of 99, we are left with - \(\frac{17}{99}\)
Re: What is the remainder when 2^99 is divided by 99?
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12 Nov 2019, 12:30
Could you please elaborate on this "rule" further? I tried it with different numbers, and I was not able to obtain the same result.
mandeepkathuria wrote:
Bunuel wrote:
What is the remainder when \(2^{99}\) is divided by 99?
A. 17 B. 15 C. 13 D. 11 E. 9
since 99 can be written as 3*3*11, and none of the three prime factors can be cancelled with the numerator. we simply add the three primes. (3+3+11 = 17) answer - A) 17
Re: What is the remainder when 2^99 is divided by 99?
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26 Dec 2019, 12:31
nick1816 wrote:
8=1*9-1
\(2^3=-1 mod 9\) \((2^3)^{33}=(-1)^{33}\) mod 9
\(2^{99}\)= -1 mod 9= 8 mod 9
\(2^{99}=9m+8\).......(1)
32=3*11-1 \(2^5\)=-1 mod 11 \((2^5)^{19}\)=-1 mod 11 \(2^{95}*2^4\)=-1*5 mod 11 \(2^{99}\)= -5 mod 11= (11-5) mod 11 \(2^{99}\)= 6 mod 11
\(2^{99}=11n+6\).......(2)
So basically our question stem is "The remainder when N is divided by 9 is 8, and when divided by 11 is 6. What is the remainder when N is divided by 99"?
from 1 and 2
\(2^{99}= lcm(9,11)x+17\) \(2^{99}= 99x+17\)
Bunuel wrote:
What is the remainder when \(2^{99}\) is divided by 99?
Re: What is the remainder when 2^99 is divided by 99?
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26 Dec 2019, 14:30
Yup. i split 99 into 9 and 11 because 9 and 11 are co-prime; hence lcm(9,11)=99
M*N= x mod y
M= a mod y N= b mod y
M*N= (a*b) mod y, where a*b<y
Then x= a*b
ShankSouljaBoi wrote:
nick1816 wrote:
8=1*9-1
\(2^3=-1 mod 9\) \((2^3)^{33}=(-1)^{33}\) mod 9
\(2^{99}\)= -1 mod 9= 8 mod 9
\(2^{99}=9m+8\).......(1)
32=3*11-1 \(2^5\)=-1 mod 11 \((2^5)^{19}\)=-1 mod 11 \(2^{95}*2^4\)=-1*5 mod 11 \(2^{99}\)= -5 mod 11= (11-5) mod 11 \(2^{99}\)= 6 mod 11
\(2^{99}=11n+6\).......(2)
So basically our question stem is "The remainder when N is divided by 9 is 8, and when divided by 11 is 6. What is the remainder when N is divided by 99"?
from 1 and 2
\(2^{99}= lcm(9,11)x+17\) \(2^{99}= 99x+17\)
Bunuel wrote:
What is the remainder when \(2^{99}\) is divided by 99?
I would find the reminders on division by 9 and 11 separately first.
\(2^{99} = 8^33 = (9 - 1)^33\) On division by 9, it will leave remainder (-1) which is same as remainder 8.
\(2^{99} = 2^4 * 2^{95} = 16 * (32)^{19} = 16 * (33 - 1)^19\) On division by 11, it will leave remainder -16 which is same as -5 which is same as 6.
So the number upon division by 9 leaves remainder 8 and upon division by 11 leaves remainder 6. This is now our remainders question. The first such number will be 17. So upon division by 99, remainder will be 17.
Answer (A)
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I would find the reminders on division by 9 and 11 separately first.
\(2^{99} = 8^33 = (9 - 1)^33\) On division by 9, it will leave remainder (-1) which is same as remainder 8.
\(2^{99} = 2^4 * 2^{95} = 16 * (32)^{19} = 16 * (33 - 1)^19\) On division by 11, it will leave remainder -16 which is same as -5 which is same as 6.
So the number upon division by 9 leaves remainder 8 and upon division by 11 leaves remainder 6. This is now our remainders question. The first such number will be 17. So upon division by 99, remainder will be 17.
Answer (A)
Quote:
I’ve been struggling to understand the solution you have given here. I love to learn from you much because your solutions usually are easy to grasp. Would it be possible to get a bit more detailed explanation from you of this question?
Yes, the solution here is a lot less intuitive and a lot more 'mathematical'. This happens when we try to practice for GMAT using non-GMAT sources. Tough GMAT questions are usually very interesting and fun in a way. I have made mistakes in a few of them even after so many years and that is what is challenging about them - you can never fully prepare for them because they have a punch. Of course, you will do well on most of them if you have your concepts sorted so scoring 51 will not be problem but still every now and then you will get something that will knock you off your feet. But I digress. Coming back to this question, I could see that no power of 2 that I knew is very close to a multiple of 100. On the other hand, I notice that powers 8 and 32 are 1 less than multiples of 9 and 11. Then perhaps that is where the answer lies.
Using binomial theorem I discussed in this post (https://www.veritasprep.com/blog/2011/0 ... ek-in-you/), we know that When 2^99 is divided by 9, it will leave remainder 8. When 2^99 is divided by 11, it will leave remainder 6.
Now, isn't it similar to questions like this: When n is divided by 9, it leaves remainder 8 and when it is divided by 11, it leaves remainder 6. What is the remainder when n is divided by 99?
I would find the reminders on division by 9 and 11 separately first.
\(2^{99} = 8^33 = (9 - 1)^33\) On division by 9, it will leave remainder (-1) which is same as remainder 8.
\(2^{99} = 2^4 * 2^{95} = 16 * (32)^{19} = 16 * (33 - 1)^19\) On division by 11, it will leave remainder -16 which is same as -5 which is same as 6.
So the number upon division by 9 leaves remainder 8 and upon division by 11 leaves remainder 6. This is now our remainders question. The first such number will be 17. So upon division by 99, remainder will be 17.
Answer (A)
Quote:
I’ve been struggling to understand the solution you have given here. I love to learn from you much because your solutions usually are easy to grasp. Would it be possible to get a bit more detailed explanation from you of this question?
Yes, the solution here is a lot less intuitive and a lot more 'mathematical'. This happens when we try to practice for GMAT using non-GMAT sources. Tough GMAT questions are usually very interesting and fun in a way. I have made mistakes in a few of them even after so many years and that is what is challenging about them - you can never fully prepare for them because they have a punch. Of course, you will do well on most of them if you have your concepts sorted so scoring 51 will not be problem but still every now and then you will get something that will knock you off your feet. But I digress. Coming back to this question, I could see that no power of 2 that I knew is very close to a multiple of 100. On the other hand, I notice that powers 8 and 32 are 1 less than multiples of 9 and 11. Then perhaps that is where the answer lies.
Using binomial theorem I discussed in this post (https://www.veritasprep.com/blog/2011/0 ... ek-in-you/), we know that When 2^99 is divided by 9, it will leave remainder 8. When 2^99 is divided by 11, it will leave remainder 6.
Now, isn't it similar to questions like this: When n is divided by 9, it leaves remainder 8 and when it is divided by 11, it leaves remainder 6. What is the remainder when n is divided by 99?
n = 9a + 8 n = 11b + 6 Try b = 1, n = 17. It is of the form 9a + 8 too so first such number is 17.
n = 99c + 17
Then, on division by 99, the remainder will be 17.
Wow. You were so quick with the response. Thank you very much for being so approachable even on holidays. That’s true commitment. Once again, wish you great holidays!
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