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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]

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New post 27 Nov 2007, 14:38
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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

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New post 27 Nov 2007, 14:57
6
2
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]

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New post 27 Nov 2007, 15:37
1
1
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]

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New post 27 Nov 2007, 22:57
2
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]

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New post 27 Nov 2007, 23:15
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]

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New post 14 Nov 2010, 23:40
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]

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New post 15 Nov 2010, 00:27
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]

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New post 18 Jan 2012, 12:54
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]

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New post 10 Aug 2013, 00:44
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 becomes\(7^3\)

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]

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New post 10 Aug 2013, 02:39
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]

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New post 06 Apr 2014, 07:15
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]

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New post 06 Apr 2014, 07:38
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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?
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.

E.
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Re: If n is a positive integer, what is the remainder when  [#permalink]

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New post 07 Apr 2014, 19:55
\(7^{4n+3} * 6^n\)

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]

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New post 11 Sep 2017, 16:01
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


Simplifying the expression, we have:

(7^4n)(7^3)(6^n)

We need to determine the remainder when the expression above is divided by 10; the remainder will be equal to the units digit of that expression.

Since 6^n will always have a units digit of 6, let’s determine the units digits of (7^4n) and (7^3):

Let’s examine the units digits of 7^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 7. When writing out the pattern, notice that we are concerned ONLY with the units digit of 7 raised to a power.

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

7^5 = 7

The pattern of the units digits of powers of 7 repeats every 4 exponents. The pattern is 7–9–3–1. In this pattern, all positive exponents that are multiples of 4 will produce 1 as its units digit. Thus:

7^3 has a units digit of 3 and 7^4n has a units digit of 1.

Since 1 x 3 x 6 = 18, (7^4n)(7^3)(6^n) has a units digit of 8, which is also the remainder when (7^4n)(7^3)(6^n) is divided by 10.

Answer: E
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Re: If n is a positive integer, what is the remainder when  [#permalink]

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New post 16 May 2018, 04:21
what is the unit digit of (5^n +4^2n+7^4n)^4n
Re: If n is a positive integer, what is the remainder when &nbs [#permalink] 16 May 2018, 04:21
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