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If you divide 7^131 by 5, which remainder do you get? [#permalink]
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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
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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]
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01 Sep 2009, 04:18
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Same as asking If you divide 7^3 by 5, which remainder do you get?hence OA must be d) 3because 131=4(K) + 3now where from I am getting these follow the patternwhen you place 1,you get 7^1 and results in 7 in unit digit 17 249hence 9 3343hence 3 41 now its repeated 57 69 73 81 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]
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01 Sep 2009, 05:31
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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]
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01 Sep 2009, 06:00
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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]
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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
17 249hence 9 3343hence 3 41
now its repeated
57 69 73 81 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]
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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]
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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]
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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]
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01 Sep 2009, 09:27
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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]
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01 Sep 2009, 09:36
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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]
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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]
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31 Jan 2010, 07:24
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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 = 52 = 3 (since 2 obtained is negative) Therefore D
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Re: question about remainders [#permalink]
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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 lastdigitofapower70624.html#p521005
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Re: question about remainders [#permalink]
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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]
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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 = 52 = 3 (since 2 obtained is negative) Therefore D can someone plz explain me this other way? thanks. Check this first: 218ifxandyarepositiveintegerswhatistheremainder109636.html#p875157If 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]
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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]
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Re: question about remainders [#permalink]
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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



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Re: question about remainders [#permalink]
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08 Oct 2013, 12:20
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 = 52 = 3 (since 2 obtained is negative) Therefore D Hey buddy. Thanks for the explanation. Looks cool, but could you explain what is that Mod stuff? Thanks a lot, Cheers J



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Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]
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Re: If you divide 7^131 by 5, which remainder do you get?
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