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Remainder problem

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Re: Remainder problem

by mohitsharda » Sat Sep 26, 2009 6:50 am
uptowngirl92 wrote:What is the remainder when 2^92 / 7 ?
2^1 gives remainder 2
2^2 => rem= 4
2^3 => rem= 1
2^4 => rem= 2

So, it repeats after every third power.
powers 1,4,7,10 i.e. 3k + 1 => give remainder 2
similarly 3k + 2 powers give remainder 4
92 = 3(30)+2
So, remainder is 4.
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by ssmiles08 » Sat Sep 26, 2009 7:55 am
2^92 = (2^10)^9 * 2^4

if you know your powers of 2; you know that 2^10 is 1024

1024 gives you a remainder of 2 when divided by 7

2^9 = 512

512/7 gives you a remainder of 1

2^2 gives you a remainder of 4

1*4 = 4

remainder = 4.
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by uptowngirl92 » Sat Sep 26, 2009 6:55 pm
I am MAJORLY confused..okay lets start from the beginning..

I had come across this question:
What is the remainder: 7^548/10
We know 7 has cycles of 4: 7,9,3,1,7,9,3,...
So,548=136 x 4 +4.Therefore remainder is 1.Correct ans.

Amother question:
What is the remainder: 7^131/5
Again the cycles theory,and we get the remainder as 3.Correct ans.

Now,coming back to our question:
2^92/7

Cycles theory of 2: 2,4,8,6,2,4,8...
In this way the remainder should come 6.

So the primary question is why is the third question different from the first two??Why are we not proceeding in the same way? :shock:
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by uptowngirl92 » Mon Sep 28, 2009 6:57 pm
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by xcusemeplz2009 » Tue Sep 29, 2009 7:51 am
uptowngirl92 wrote:I am MAJORLY confused..okay lets start from the beginning..

I had come across this question:
What is the remainder: 7^548/10
We know 7 has cycles of 4: 7,9,3,1,7,9,3,...
So,548=136 x 4 +4.Therefore remainder is 1.Correct ans.

Amother question:
What is the remainder: 7^131/5
Again the cycles theory,and we get the remainder as 3.Correct ans.

Now,coming back to our question:
2^92/7

Cycles theory of 2: 2,4,8,6,2,4,8...( its cycle of unit digit we need to check cycle of rem pls c below for expln.)
In this way the remainder should come 6.

So the primary question is why is the third question different from the first two??Why are we not proceeding in the same way? :shock:
i think u are confused between unit digit and remainder

for the 1st one 7^548/10

we need to check the when remainder is repeating
7^1=7 rem=7 (when divided by 10)
7^2=49 rem=9 (----do-----)
7^3=343 rem=3(----do-----)
7^4=2401 rem=1(--do--)
repeats in 4th power
so 548=4k+4 i.e to the power 4 i.e rem is 1

similarly 7^131/5
131=4k+3 i.e rem is 3

coming to original q
2^92/7
2^1=2 rem=2(when div by 7)
2^2=4 rem=4(---do--)
2^3=8 rem=1(--do---)
2^4=16 rem=2(---do---)

repetion occurs after third power therefore
92=3k+2 i.e we need to check the rem for 2^2 which is 4 hence 4 is correct
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