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Re: PS - remaineder of (7^(4n+3))(6^n) is divided by 10 [#permalink]
27 Nov 2007, 14:37

gregspirited wrote:

If n is a positive integer, what is the remainder When (7^(4n+3))(6^n) is divided by 10? F. 1 G. 2 H. 4 I. 6 J. 8

This one took me bout 3 1/2 min. Just testin numbers and what not.

First notice that n is positive. Save time by noticing that I worked out one solution where n=0 only to find that thats not an option :p.
1-7 stands for ^1 thru 7
1: 7*1=7
2: 7*7=9
3: 7*9=3
4: 7*3=1
5: 7*1=7
6: 7*7=9
7: 7*9=3

Pattern repeats every @ 5. Notice every ^4 or multiple of 4 is always going to be 1. This is just for future notice for similar problems.

so 7^4n+3 ---> if n=1 then its ((7^7)*6))/10 which can say is going to be 3*8--> 18/10 --> R=8

Now from here if id double check just to make sure.

7^4(2)+3*6^2 ---> 7^11*36 or we can just say again 7^11*6 (b/c we are only interested in the units digit).

Since ^12 is going to be 1 that means ^11 is going to be 3 (as taken from our pattern)

Re: PS - remaineder of (7^(4n+3))(6^n) is divided by 10 [#permalink]
27 Nov 2007, 21:57

2

This post received KUDOS

gregspirited wrote:

If n is a positive integer, what is the remainder When (7^(4n+3))(6^n) is divided by 10? F. 1 G. 2 H. 4 I. 6 J. 8

J

All we need to do to find the remainder is find out the units digit of the expression.

7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1

If n = 1 then 7^(4n+3) = 7^7 = 7^4*7^3
units digit of 7^4 which is 1 * units digits of 7^3 which is 3 = 3
6 raised to any non 0 positive power will have units digit of 6
therefore units digit of expression = 6*3 = 8
when divided by 10 this will always leave a remainder of 8.

Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Re: If n is a positive integer, what is the remainder When [#permalink]
18 Jan 2012, 11:54

Cyclicity of 7 is 4. if n = 1, power of 7 will 7 and, the reminder will 3 if 7 is divided by 4 if n = 3, power of 7 will 15 and, the reminder will 3 if 15 is divided by 4 Last digit of 7^4 will 3, and last digit 6 power anything will 6 Now, 6*3 = 18/10 = 8 Reminder. Ans. E _________________

Re: remainder when ((7^(4n+3)(6^n)) [#permalink]
10 Aug 2013, 01:39

monirjewel wrote:

If n is a positive integer, what is the remainder when ((7^(4n+3)(6^n)) is divided by 10? (A) 1 (B) 2 (C) 4 (D) 6 (E) 8

.................. 7^4n × 7^3 × 6^n = (7^2)2n × 343 × 6^n =(50-1)^2n × 343 × 6^n so the last term = (-1)^2n × 343 × 6^n = 343 × 6^n For any values of n 6^n = something 6 in the unit digit, and 343 × something 6 in the unit digit will always provide something 8 in the unit digit, so Answer is E _________________

Re: If n is a positive integer, what is the remainder when [#permalink]
06 Apr 2014, 06:38

Expert's post

gregspirited wrote:

If n is a positive integer, what is the remainder when (7^(4n+3))(6^n) is divided by 10? A. 1 B. 2 C. 4 D. 6 E. 8

Theory : The cyclicity for the unit's digit for 7 repeats at an interval of 4. Thus, units digit for \(7^1 = 7, 7^2 = 9 , 7^3 = 3\) and \(7^4 = 1\)

Given expression : \(7^{4n}*7^{3}*6^n\) and note that n is a positive integer.

As 4n is always a multiple of 4, the units digit of \(7^{4n}\) will always be 1. Units digit of\(7^3 = 3\). Also, \(6^n\) will always have the same units digit of 6, just as\(5^n\)(units digit of 5) does.

Thus, final expression will have the unit's digit as : \(1*3*6 = 18\). As the divisor is 10, the remainder will always be the units digit = 8.

Re: If n is a positive integer, what is the remainder when [#permalink]
08 Apr 2015, 09:52

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