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ok i calculated the answer as 8 but its too complex to be termed as "fastest" way
Yeah its 8... I know the method but its very long....
1. Refine it: 49 = 23*2+3. 3 is the remainder of 49 divided by 23 => 49^1000 has the same remainder, when divided by 23, as 3^1000!
2. Find the CYCLE:
Start with small powers of 3 and find remainders:
3^0 = 1. Remainder = 1.
3^1 = 3. Remainder = 3.
3^2 = 9. Remainder = 9.
3^3 = 27. Remainder = 4.
Then, you should find the remainder of 3^5 by 23. But that number is big... What to do? You should do the following: take previous remainder of 3^4 by 23 (=12), multiply it by 3, and get the new remainder! Which is, it is easy to see, 13!
To get a remainder of 3^6 by 23, you need to multiply previous remainder (=13) by 3 and get the new remainder: 39. This number is greater that 23. So, divide it by 23 again, and get the new remainder: 16!
3^4 = 81. Remainder = 12.
3^5 = 243. Remainder = 13.
3^6 = ...{something you can't calculate}. Remainder = 13*3 mod 23 = 16!
3^7 = ...{something you can't calculate}. Remainder = 16*3 mod 23 = 2!
So, 3^11 mod 23 = 1. Then, it is not surprising, 3^12 mod 23 = 3... and the CYCLE starts AGAIN.
What you need to do at this stage, is to find a minimum duration of CYCLE.
In our case, duration of CYCLE is 11 - 0 = 11.
3. Once you have found the PERIOD of the CYCLE, you can easily compute the remainder of 3^1000 mod 23. 1000 = 11{=PERIOD}*90 + 10. => the remainder of 1000 by PERIOD{=11} is 10! => 49^1000 mod 23 = 3^1000 mod 23 = 3^10 mod 23 = 8, which was previously calculated!
Re: Remainder: quickest way...tough one [#permalink]
22 Aug 2007, 16:12
fresinha12 wrote:
what is the remainder of 49^1000 divided by 23?
1) 9 2) 8 3) 7 4) 6 5) 1
I am not sure if you will see something like this in GMAT either, but here it goes my fastest method...
49^1000 / 23, you know that 23*2=46; thus, use the remainder of 3 and replace 49 with 3, you get:
3^1000 / 23
Now, work out some numbers...you know that 3^5 = 81*3 = 243. If 243/23, you get remainder of 13 since 243-230 = 13. Plug that back in just like the one above, you get something like this:
((3^5)^200) / 23 => plug in => (13^200)/23 = ((13^2)^100)/23
You know that 13^2 = 26, so it will have remainder of 3 when divide by 26. Plug in back in the same way...
3^100 / 23
Keep using the same method, you get:
3^100 / 23 = ((3^5)^20)/23 => (13^20)/23 = ((13^2)^10)/23
=> (3^10) / 23 = ((3^5)^2) / 23 => (13^2) / 23 = 169/23
Find the remainder for the final number: 23*7 = 161
169-161 = 8
ok i calculated the answer as 8 but its too complex to be termed as "fastest" way
Yeah its 8... I know the method but its very long....
1. Refine it: 49 = 23*2+3. 3 is the remainder of 49 divided by 23 => 49^1000 has the same remainder, when divided by 23, as 3^1000!
2. Find the CYCLE:
Start with small powers of 3 and find remainders:
3^0 = 1. Remainder = 1.
3^1 = 3. Remainder = 3.
3^2 = 9. Remainder = 9.
3^3 = 27. Remainder = 4.
Then, you should find the remainder of 3^5 by 23. But that number is big... What to do? You should do the following: take previous remainder of 3^4 by 23 (=12), multiply it by 3, and get the new remainder! Which is, it is easy to see, 13!
To get a remainder of 3^6 by 23, you need to multiply previous remainder (=13) by 3 and get the new remainder: 39. This number is greater that 23. So, divide it by 23 again, and get the new remainder: 16!
3^4 = 81. Remainder = 12.
3^5 = 243. Remainder = 13.
3^6 = ...{something you can't calculate}. Remainder = 13*3 mod 23 = 16!
3^7 = ...{something you can't calculate}. Remainder = 16*3 mod 23 = 2!
So, 3^11 mod 23 = 1. Then, it is not surprising, 3^12 mod 23 = 3... and the CYCLE starts AGAIN.
What you need to do at this stage, is to find a minimum duration of CYCLE.
In our case, duration of CYCLE is 11 - 0 = 11.
3. Once you have found the PERIOD of the CYCLE, you can easily compute the remainder of 3^1000 mod 23. 1000 = 11{=PERIOD}*90 + 10. => the remainder of 1000 by PERIOD{=11} is 10! => 49^1000 mod 23 = 3^1000 mod 23 = 3^10 mod 23 = 8, which was previously calculated!
nice copy and paste....
just kidding...
but yeah this is a lengthy approach...best would be make an educated guess.
Re: Remainder: quickest way...tough one [#permalink]
11 Jan 2008, 07:59
kazakhb wrote:
marcodonzelli wrote:
fresinha12 wrote:
what is the remainder of 49^1000 divided by 23?
1) 9 2) 8 3) 7 4) 6 5) 1
7^2000/3.....7^2000 terminates in 7 (7,9,3,1,7....), 3*2=6; 7-6=1 OA is 5
don't follow your steps, could you explain? moreover isn't OA is 8?
I'm sorry, I didn't read 23....OA is 8:
49^1000 is a number terminating in 1. since the are no common factor b/w 49^1000 and 23, we have a number terminating in 1 divided by a prime number (23). thus, the remainder would be 8....
Re: Remainder: quickest way...tough one [#permalink]
11 Jan 2008, 08:16
Quote:
we have a number terminating in 1 divided by a prime number (23). thus, the remainder would be 8....
can someone explain this last step in more detail? Up to the part where we know 49^1000 ends in 1, but not how we know that 49^1000/23 has a remainder of 8.
Re: Remainder: quickest way...tough one [#permalink]
11 Jan 2008, 12:15
eschn3am wrote:
Quote:
we have a number terminating in 1 divided by a prime number (23). thus, the remainder would be 8....
can someone explain this last step in more detail? Up to the part where we know 49^1000 ends in 1, but not how we know that 49^1000/23 has a remainder of 8.
my reasoning is: since we have 1 as unit digit, the remainder must end in 8, which actually means 8 and 18, not 28 because 28>23. since we don't have 18 as answer choice, we can say only 8...what do you think about that?
anyway 23*104=2392 7^4=2401 remainder is 9..am i going crazy????
Re: Remainder: quickest way...tough one [#permalink]
29 Feb 2008, 08:57
49^1000 = 7^1000 * 7^1000 Using the 7,9,3,1 repeat rule, we have X1 * X1 = Y1
If Y = any digit besides 3 or 0, the remainder is larger than any of the answer choices. (lets say that Y results after doing long division by 23) If Y = 3 the remainder is 8 If Y = 0 the remainder is 1 This could be a quick 50/50 guess, but I think the 0 possibility can be eliminated. A quick check, some number divisible by 23 (such as 230), with a 01 added at the end....23001 (since this would give a remainder of 1) Could such a number occur from multiplying 7*7*7 etc. I'm not sure, but this one isn't divisible by 7, so my guess was 8 as the remainder.
gmatclubot
Re: Remainder: quickest way...tough one
[#permalink]
29 Feb 2008, 08:57