If n is a positive integer, what is the remainder when 7^(4n+3)*6^n is

Question

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

https://gmatclub.com/forum/download/file.php?id=77024

walker · Accepted Answer

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)

GMATBLACKBELT · Answer

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) so again 3*6=18/10 ---> R =8. E or J in this problem.

GK_Gmat · Answer

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.

beckee529 · Answer

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

stanford2012 · Answer

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.

MHIKER · Answer

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

fozzzy · Answer

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

Asifpirlo · Answer

..................
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

mvmarcus7 · Answer

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

mau5 · Answer

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.

PareshGmat · Answer

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

ScottTargetTestPrep · Answer

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

CapnSal · Answer

Bunuel . This is a question from GMAT Focus Prep Mock 3, Please update the tag

Bunuel · Answer

Done. Thank you!

P.S. Can you please post the screenshot of the question? That would help. Thank you.

CapnSal · Answer

Attached the screenshot

Bunuel · Answer

Great. Thank you!

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If n is a positive integer, what is the remainder when 7^(4n+3)*6^n is

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