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If n is a positive integer, what is the remainder when

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If n is a positive integer, what is the remainder when [#permalink] New post 27 Nov 2007, 13:38
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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
[Reveal] Spoiler: OA

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 [#permalink] New post 27 Nov 2007, 13:57
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E (J). 8

last digits of 7^m
7^1 : 7
7^2 : 9
7^3 : 3
7^4 : 1
7^5 : 7
period=4 ==> last digit: 7^(4n+3) = 7^(4+3) = 3
last digit of 6^n always 6.

3*6 ==> 8 - reminder (last digit)
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Re: PS - remaineder of (7^(4n+3))(6^n) is divided by 10 [#permalink] New post 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 :P 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)

so again 3*6=18/10 ---> R =8.


E or J in this problem.
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Re: PS - remaineder of (7^(4n+3))(6^n) is divided by 10 [#permalink] New post 27 Nov 2007, 21:57
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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.
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Re: PS - remaineder of (7^(4n+3))(6^n) is divided by 10 [#permalink] New post 27 Nov 2007, 22:15
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


i got 8 after working it out in less than a minute

7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
repeats we're looking for 3rd: 3
6^n = always ends in 6
3*6 = ends in 8

when divided by 10 will leave a remainder of 8
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If n is a positive integer, what is the remainder when ((7^( [#permalink] New post 14 Nov 2010, 22:40
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
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Re: remainder when ((7^(4n+3)(6^n)) [#permalink] New post 14 Nov 2010, 23:27
Use n = 1 and plug into the equation => 7^7*6^1 / 10

Check the cyclicity of n:
7^1 = 7
7^2 = 49
7^3 = ...9*7 =...63
7^4 = ...3*7 =...21
7^5 = ...1*7 =....7 (last digit the same as for 7^1)
=> 7^7 = ....3

....3*6^1 = ....18. Since the number ends in 8 the remainder when divided by 10 must be 8.

Hence, solution E is correct.
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Re: If n is a positive integer, what is the remainder When [#permalink] New post 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
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Re: remainder when ((7^(4n+3)(6^n)) [#permalink] New post 09 Aug 2013, 23:44
6 and 5 are 2 numbers that have the same units digit all the time 6^6 = 6 5^5 = 5

So in this question 6^n = 6

7^{4n+3} test some values say n=0,1 etc

we get a pattern if 0 then it becomes7^3

if n = 1 then it becomes 7^7


The pattern for 7 is {7,9,3,1}

Units digit 7^{4n+3} = 3 and units digit of 6^n = 6

so 18 divided by 10 remainder is 8

Answer E
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Re: remainder when ((7^(4n+3)(6^n)) [#permalink] New post 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
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 06 Apr 2014, 06:15
Answer is E = 8

((7^(4n+3))*6^n)/10

Trying for 1 we have:

((7^7)*(6^1))/10 = (7*6*(7^6))/10 = (42*(7^6))/10 = 4.2*(7^6).

7^2 = 9
7^3 = 9*7 = 3
7^4 = 3*7 = 1
7^5 = 1*7 = 7
7^6 = 7*7 = 9

9*2"from 4.2" gives us 8
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 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 of7^3 = 3. Also, 6^n will always have the same units digit of 6, just as5^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.

E.
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Re: If n is a positive integer, what is the remainder when [#permalink] New post 07 Apr 2014, 18:55
7^{4n+3} * 6^n

Placing value of n=1

= 7^7 * 6

= 7^7 (10 - 4)

= 7^7 * 10 - 7^7 * 4

7^7 * 10 >> Will not leave any remainder

- 7^7 * 4 >> Cyclicity for power 7 = 7, 9, 3, 1

7^7 gives 3 in the units place & multiplying by 4 gives 2 in units place

So 10 - 2 = 8

Answer = E
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Re: If n is a positive integer, what is the remainder when   [#permalink] 07 Apr 2014, 18:55
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