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If you divide 7^131 by 5, which remainder do you get?

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If you divide 7^131 by 5, which remainder do you get? [#permalink] New post 01 Sep 2009, 04:06
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If you divide 7^131 by 5, which remainder do you get?

A. 0
B. 1
C. 2
D. 3
E. 4
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Oct 2012, 03:39, edited 1 time in total.
Renamed the topic and edited the question.
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Re: question about remainders [#permalink] New post 01 Sep 2009, 04:18
Same as asking

If you divide 7^3 by 5, which remainder do you get?
hence OA must be d) 3

because 131=4(K) + 3

now where from I am getting these
follow the pattern

when you place 1,you get 7^1 and results in 7 in unit digit

1---7
2---49--hence 9
3---343---hence 3
4--1


now its repeated


5-7
6-9
7-3
8-1
so on..

131=128+3=4(32)+3

hope i am making sense:)


blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?

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Re: question about remainders [#permalink] New post 01 Sep 2009, 05:31
whenever you come across a question like this :-- a number raised to higher power of integer divided by 2,3,4,5,6,7,8,9..
and asking for remainder...then, the question is actually asking for the unit digit of the numerator..

now,
3^4 has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 3.

6 to the power anything, always has a unit digit of 6

7^4 also has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 7

now, 151 = 148 + 3

7^148 * 7^3 = unit digit of 1 * unit digit of 3 = unit digit of 3..

any number divided by 5 gives remainder = unit digit of the number, when unit digit is less than 5

and (unit digit - 5) , when unit digit is greator than 5
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Re: question about remainders [#permalink] New post 01 Sep 2009, 06:00
gmate2010 wrote:
whenever you come across a question like this :-- a number raised to higher power of integer divided by 2,3,4,5,6,7,8,9..
and asking for remainder...then, the question is actually asking for the unit digit of the numerator..

now,
3^4 has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 3.

6 to the power anything, always has a unit digit of 6

7^4 also has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 7

now, 151 = 148 + 3

7^148 * 7^3 = unit digit of 1 * unit digit of 3 = unit digit of 3..

any number divided by 5 gives remainder = unit digit of the number, when unit digit is less than 5

and (unit digit - 5) , when unit digit is greator than 5


Gmate2010, I don't get it.. why do we consider 151?.. And how do we conclude that 151 is necessarily X+3??? Because we know that 1^3 is equal to 3? I sound like a diletant...)
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Re: question about remainders [#permalink] New post 01 Sep 2009, 06:06
nitya34 wrote:
Same as asking

If you divide 7^3 by 5, which remainder do you get?
hence OA must be d) 3

because 131=4(K) + 3

now where from I am getting these
follow the pattern

when you place 1,you get 7^1 and results in 7 in unit digit

1---7
2---49--hence 9
3---343---hence 3
4--1


now its repeated


5-7
6-9
7-3
8-1
so on..

131=128+3=4(32)+3

hope i am making sense:)


No, truely, I don't understand. How do we know that 131 is exactly 4K+3.. Beats me!!!!!!!!!!!!!!!!!!!! I see the vicious circle of powers but don't get the conclusion...
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Re: question about remainders [#permalink] New post 01 Sep 2009, 06:43
let me try again...
we have to find the remainder of the [7^131 by 5]


Now 7^131 is XYZ....XXXXXXXX...XXXXX(3)[Note the earlier post of mine...]
where X ,Y,Z are unknown Integers

hence (7^131) divided by 5 will yield a remainder of 3
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Last edited by nitya34 on 01 Sep 2009, 06:51, edited 2 times in total.
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Re: question about remainders [#permalink] New post 01 Sep 2009, 06:47
CasperMonday wrote:
gmate2010 wrote:
whenever you come across a question like this :-- a number raised to higher power of integer divided by 2,3,4,5,6,7,8,9..
and asking for remainder...then, the question is actually asking for the unit digit of the numerator..

now,
3^4 has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 3.

6 to the power anything, always has a unit digit of 6

7^4 also has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 7

now, 151 = 148 + 3

7^148 * 7^3 = unit digit of 1 * unit digit of 3 = unit digit of 3..

any number divided by 5 gives remainder = unit digit of the number, when unit digit is less than 5

and (unit digit - 5) , when unit digit is greator than 5


Gmate2010, I don't get it.. why do we consider 151?.. And how do we conclude that 151 is necessarily X+3??? Because we know that 1^3 is equal to 3? I sound like a diletant...)


Sorry, its 131..

if we divide 131 by 4 then we get remainder 3..

=> 7 raised to the power factor of 4 will have a unit digit 1.. + 7*3 will have a unit digit 3
=> unit digit 1 * unit digt 3 = unit digit 3
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Re: question about remainders [#permalink] New post 01 Sep 2009, 08:50
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?


If you look at the power of 7, it shows a repeated trend at the Unit's digit. For e.g.:
7^1 = 7 (7 at unit place)
7^2 = 49 (9 at unit place)
7^3 = 343 (3 at unit place)
7^4 = 2401 (1 at unit place)
7^5 = 16807 (7 at unit place)

So if you see, this trend of 7,9,3,1......7,9,3,1.......repeats itself.
7^131 will give 3 as unit's digit and when it is divided by 5, it will give the remainder as 3.

Hope it helps.
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Re: question about remainders [#permalink] New post 01 Sep 2009, 09:27
when you divide 7 by 5 we get 2 as remainder

so now we can write it as 7^131 \ 5 = 2 ^131 \ 5

= 2^ 120 * 2^ 11 \ 5
= (2^4)^30 * 2048 ( when we multiply 2^11 we will get the value as 2048)
= 16^30 * 2048
= 1^ 30 * 2048 ( when we divide 16 by 5 the remainder is 1 so 1^30 = 1

finally divide 2048 by 5 and u get the remainder = 3

hence 1 * 3 = 3

So answer is D
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Re: question about remainders [#permalink] New post 01 Sep 2009, 09:36
there is another easy method

cyclicity of 7 = 4

so when we divide 131 by 4 we get 3 as remainder

therefore 7^3 = 343

now divide 343 by 5 and u get the remainder as 3

so answer is D........

Note : cyclicity is used to find the last digit of the number i.e in case of powers it is specially usefull
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Re: question about remainders [#permalink] New post 01 Sep 2009, 20:19
7*7*7*7= xxx1 -> only last digit is important.

you can use this for 32 time and get to 7^128 and last digit will be 1.

1 *7*7*7 = 343 and divided by 5 leaves a remainder of 3.
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Re: question about remainders [#permalink] New post 31 Jan 2010, 07:24
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?


7^131 Mod 5
= (7 Mod 5) ^ 131
= 2 ^ 131
= (2 Mod 5)^ 130 * (2 Mod 5)
= (4 Mod 5)^65 * 2
= (-1)^65 * 2
= -2

Therefore remainder = 5-2 = 3 (since -2 obtained is negative)

Therefore D
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Re: question about remainders [#permalink] New post 12 Feb 2010, 09:20
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?


7^131 / 5 and have to find the remainder.

as per the theory, 7 has the cyclicity of 4. Therefore, 131 mod 4 = 3 so unit digit of 7^131 will be same as 7^3 = 343

343/5 gives 3 as remainder.

D







For Details please check this thread
last-digit-of-a-power-70624.html#p521005
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Re: question about remainders [#permalink] New post 07 Dec 2010, 11:18
7^1 = 7 => 7/5 has R2
7^2 = 49 => 49/5 has R4
7^3 = 343 => 343/5 has R3
7^4 = 2401 =>2401/5 has R1
7^5 = 16807 => 16807/5 has R2

So it repeats every 4. 131/4 = 128 R3
So 7^128 / 5 has the same remainder as 7^4 /5
7^129 / 5 has the same remainder as 7^1 / 5
7^130 / 5 has the same remainder as 7^2 / 5
7^131 / 5 has the same remainder as 7^3 / 5, which is 3.

The answer is 3.
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Re: question about remainders [#permalink] New post 19 Feb 2011, 03:26
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144144 wrote:
jeeteshsingh wrote:
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?


7^131 Mod 5
= (7 Mod 5) ^ 131
= 2 ^ 131
= (2 Mod 5)^ 130 * (2 Mod 5)
= (4 Mod 5)^65 * 2
= (-1)^65 * 2
= -2

Therefore remainder = 5-2 = 3 (since -2 obtained is negative)

Therefore D


can someone plz explain me this other way? thanks.


Check this first: 218-if-x-and-y-are-positive-integers-what-is-the-remainder-109636.html#p875157

If you divide 7^131 by 5, which remainder do you get?
A. 0
B. 1
C. 2
D. 3
E. 4

Last digit of 7^(positive integer) repeats in blocks of 4: {7, 9, 3, 1} - {7, 9, 3, 1} - ... (cyclicity of 7 in power is 4).

As the remainder upon division 131 by 4 (cyclicity) is 3 then the last digit of 7^131 is the same as that of 7^3 so 3 (the third digit from the pattern {7, 9, 3, 1}). Now, any positive integer ending with 3 upon division by 5 yields the remainder of 3.

Answer: D.

Hope it's clear.
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Re: question about remainders [#permalink] New post 19 Feb 2011, 03:34
i understand ur way perfectly and this is the way i solved it as well, but im trying to understand the solution of blackcrow.

he divided 7/5 and left with 2... i understand some of the logic but im not sure how to put it on paper and solve it the way he is doing it.
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Re: question about remainders [#permalink] New post 19 Feb 2011, 03:35
anyways forgot +1. given. thanks.
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Re: question about remainders [#permalink] New post 19 Oct 2012, 03:37
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?


For finding remainders by 5 & 10 : find the last digit

7 has a cyclicity of 4 . : 7,9,3,1

7^131

So 131/4 .. last digit = 3
so the remainder is 3

Answer e
Re: question about remainders   [#permalink] 19 Oct 2012, 03:37
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