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Re: question about remainders [#permalink]
01 Sep 2009, 04:31
1
This post received KUDOS
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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
Re: question about remainders [#permalink]
01 Sep 2009, 05:00
1
This post was BOOKMARKED
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...)
Re: question about remainders [#permalink]
01 Sep 2009, 05:06
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...
Re: question about remainders [#permalink]
01 Sep 2009, 05:47
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
Re: question about remainders [#permalink]
01 Sep 2009, 07:50
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.
Re: question about remainders [#permalink]
01 Sep 2009, 08:27
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
Cheers! JT........... If u like my post..... payback in Kudos!!
|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|
Re: question about remainders [#permalink]
07 Dec 2010, 10:18
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.
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.
Re: question about remainders [#permalink]
19 Feb 2011, 02:34
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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