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Screw these remainders

by bblast » Wed Apr 27, 2011 9:56 am
What is the remainder when 51^203 is divided by 7?

a>4
b>2
c>1
d>6
e>5

oa-4


What is the remainder when 2^1000 is divided by 3?
1>1
b>2
c>4
d>6
e>0

oa-a
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by manpsingh87 » Wed Apr 27, 2011 10:05 am
bblast wrote:What is the remainder when 51^203 is divided by 7?

a>4
b>2
c>1
d>6
e>5

oa-4


What is the remainder when 2^1000 is divided by 3?
1>1
b>2
c>4
d>6
e>0

oa-a
51^203; 51 when divided by 7; gives 2 as a remainder hence expression 51^203 can be reduced to 2^203;
2^3=8 8 when divided by 7 gives 1 as remainder; now divide 203/3 we get 2 as remainder;
hence our expression is (8^67) * 2^2/7; which gives 4 as a remainder hence A

Q2. 2^2/3 gives 1 as remainder; hence our expression will reduce to 4^500/3; which will give 1 as a remainder..!! hence A
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by MAAJ » Wed Apr 27, 2011 1:12 pm
Brilliant! Learned something new, thanks!
manpsingh87 wrote:
bblast wrote:What is the remainder when 51^203 is divided by 7?

a>4
b>2
c>1
d>6
e>5

oa-4


What is the remainder when 2^1000 is divided by 3?
1>1
b>2
c>4
d>6
e>0

oa-a
51^203; 51 when divided by 7; gives 2 as a remainder hence expression 51^203 can be reduced to 2^203;
2^3=8 8 when divided by 7 gives 1 as remainder; now divide 203/3 we get 2 as remainder;
hence our expression is (8^67) * 2^2/7; which gives 4 as a remainder hence A

Q2. 2^2/3 gives 1 as remainder; hence our expression will reduce to 4^500/3; which will give 1 as a remainder..!! hence A
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by pemdas » Wed Apr 27, 2011 1:46 pm
I cannot agree with the following
manpsingh87 wrote:51^203; 51 when divided by 7; gives 2 as a remainder hence expression 51^203 can be reduced to 2^203; 2^3=8 8 when divided by 7 gives 1 as remainder; now divide 203/3 we get 2 as remainder; hence our expression is (8^67) * 2^2/7; which gives 4 as a remainder hence A
51^203 : 7 = (7*7 + 2)^203 : 7. Can we divide (a+b)^n : c to get b^n?

here's one approach to this problem -> last two digits always repeated when 51^odd and 51^even (example, 51^2 last two digits are 01, 51^3 last two digits are 51). Since the power 203 is odd we take 51^odd and the last two digits are 51. The third digit from the end is hundredth digit. The hundredth digits divided by 7 may give various remainders. The pattern for the hundredth digits is such that for every 7th power the hundredth digit is 8. Hence 851 divided by 7 will give the remainder of 4.
What is the remainder when 2^1000 is divided by 3?
the obvious approach is 2^(10*10*10) or 2^10^3. Since 2^10 is 1024, the expression 1024^100 has the last digit 6 and the remainder of either 0 or 1. We decide about having the remainder of 1, as all our answer choices contain the remainders e) >0, a)>1 etc. So the remainder must be the tenth digit divided by 3 and giving the remainder of 1.
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by Ian Stewart » Wed Apr 27, 2011 4:36 pm
pemdas wrote:I cannot agree with the following
manpsingh87 wrote:51^203; 51 when divided by 7; gives 2 as a remainder hence expression 51^203 can be reduced to 2^203; 2^3=8 8 when divided by 7 gives 1 as remainder; now divide 203/3 we get 2 as remainder; hence our expression is (8^67) * 2^2/7; which gives 4 as a remainder hence A
51^203 : 7 = (7*7 + 2)^203 : 7. Can we divide (a+b)^n : c to get b^n?
The questions in the original post are not at all in the style of real GMAT remainders questions, so no one here needs to be concerned about how to solve them, but manpsingh's work was perfectly correct. If you are asked for the remainder when 51^203 is divided by 7, then you can replace the 51 with 2 if you like, since 51 gives a remainder of 2 when divided by 7.

The reason is the following: if you take any number at all with a remainder of 2 when divided by 7, then your number is in the form 7q + 2. If you were to then expand (7q + 2)^203 completely, then with the lone exception of the 2^203 term, every single term in that expansion will be a multiple of 7, and thus won't have any influence on the remainder. It is probably easier to see why this is true by looking at a much smaller power than 203. If you take, say:

(7q + 2)^2 = (7q)^2 + (2)(2)(7q) + 2^2

every term is a multiple of 7 besides the '2^2', so the remainder will always be 4 when (7q + 2)^2 is divided by 7. That is, when you divide something like 51^2 (or 58^2, or 702^2, etc) by 7, the remainder will certainly be 4.

It is, however, only very rarely useful to understand this on the GMAT.
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by manpsingh87 » Wed Apr 27, 2011 7:36 pm
pemdas wrote:I cannot agree with the following
manpsingh87 wrote:51^203; 51 when divided by 7; gives 2 as a remainder hence expression 51^203 can be reduced to 2^203; 2^3=8 8 when divided by 7 gives 1 as remainder; now divide 203/3 we get 2 as remainder; hence our expression is (8^67) * 2^2/7; which gives 4 as a remainder hence A
51^203 : 7 = (7*7 + 2)^203 : 7. Can we divide (a+b)^n : c to get b^n?

here's one approach to this problem -> last two digits always repeated when 51^odd and 51^even (example, 51^2 last two digits are 01, 51^3 last two digits are 51). Since the power 203 is odd we take 51^odd and the last two digits are 51. The third digit from the end is hundredth digit. The hundredth digits divided by 7 may give various remainders. The pattern for the hundredth digits is such that for every 7th power the hundredth digit is 8. Hence 851 divided by 7 will give the remainder of 4.
What is the remainder when 2^1000 is divided by 3?
the obvious approach is 2^(10*10*10) or 2^10^3. Since 2^10 is 1024, the expression 1024^100 has the last digit 6 and the remainder of either 0 or 1. We decide about having the remainder of 1, as all our answer choices contain the remainders e) >0, a)>1 etc. So the remainder must be the tenth digit divided by 3 and giving the remainder of 1.
well i believe Ian has already answered your query...!!! if you still have any doubt feel free to post it here,,,..!!

PS: thanks a lot Ian sir....!!!
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by pemdas » Wed Apr 27, 2011 8:14 pm
thanks Ian, while it's clear for this type of expression to reduce remainder to 2^203/7 (not analogous to 2^2 which is not divisible by 7) we are still concerned with the remainder issue, aren't we? And the fact (2^3)^67 *2^2 = 2^203, neither 8 (or 2^3) is divisible by 7 nor 2^2 is done so. Can we assess the remainder directly from 2^2/7? {(2^3)^67}/7 also leaves the remainder of 1. It's like in the following example, 2 1/5 * 3 2/5 = 11*17/25 and the remainder from 187/25 is not 2, it's 12. Can we safely assess the remainder according to explanation presented in the earlier post?
manpsingh87 wrote:...
2^3=8 8 when divided by 7 gives 1 as remainder; now divide 203/3 we get 2 as remainder;
hence our expression is (8^67) * 2^2/7; which gives 4 as a remainder hence A
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