'm' is the remainder when 3^2020 is divided by 13 and 'n' is the remai

Question

'm' is the remainder when  3^{2020}  is divided by 13 and 'n' is the remainder when  3^{2021}  is divided by 13. What is the value of m + n?

A. 6 
B. 8 
C. 10 
D. 12 
E. 14

Nups1324 · Accepted Answer

The cyclicity of the remainders when powers of 3 are divided by 13 is 3,9,1.

How?

3^1=3 >>> 3/13 leaves a remainder of 3.
3^2=9 >>> 9/13 leaves a remainder of 9.
3^3=27 >>> 27/13 leaves a remainder of 1.
3^4=81 >>> 81/13 leaves a remainder of 3. (Again)
3^5=243 >>> 243/13 leaves a remainder of 9. (Again)
And so on.

The cycle repeats after 3 numbers. So cyclicity is 3 (For Remainder)

For m the power is 2020.
2020/3 leaves a remainder of 1.

So m = 3.

For n the power is 2021.
2021/3 leaves a remainder of 2.

So n = 9.

m+n=12.

Option D. IMO.

I hope my method is correct. If not then please let me know.

Thank you.

Posted from my mobile device

ArunSharma12 · Answer

\frac{3^{2020} + 3^{2021}}{13} = \frac{3^{2020}(1 + 3) }{13} 
 \frac{27^{673}*3*4 }{13} = \frac{1^{673}*3*4 }{13} = 12 
Ans: D

VERBAL1 · Answer

Nups1324  the cyclicity of the remainders of 3 comprise 3, 9,7 (you missed this) and 1.

yashikaaggarwal · Answer

Using Euler theorem,

When A^x is divided by N, where A and N are co prime.
Since N here has only one factor other than 1
Ø(n) = n(1-1/n)
Ø(13) = 13(1-1/13)
Ø(13) = 12

when 3^2020
So A^Ø(n) mod n = 1 as remainder
3^12 mod 13 = 1
(3^12)168 mod 13 * 3^4 mod 13
1*81 mod 13

81 mod 13
3 as remainder -------------------------(1) = m

when 3^2021
So A^Ø(n) mod n = 1 as remainder
3^12 mod 13 = 1
(3^12)168 mod 13 * 3^5 mod 13
1*243 mod 13

243 mod 13
9 as remainder -----------------------------(2) = n

therefore, 
M+N = 3+9 = 12
Answer is D

Nups1324 · Answer

Hi  VERBAL1 ,

I think I'm correct. The cyclicity of powers of 3 when divided by 13.

27/13 will leave a remainder of 1.

Am I missing something?

Thank you.

Posted from my mobile device

Archit3110 · Answer

cyclicity of 3 is know
3^1 3; 3^2 9 ; 3^3 7 ; 3^4 1
since divisor is 13 so to find the remainder values start from 3^3 ,3^4,3^5 ... so on we see that the remainder follows pattern ( 1,3,9)
3^2020 would have same unit as 3^4 and remainder m would be 3 when divided by 13 and
3^2021 would have same unit as 3^5 and remainder n would be 9
m+n would be 3+9 ; 12
OPTION D

MathRevolution · Answer

Solution:

Let’s see the remainder pattern when 3 and its higher powers are divided by 13.

=> Since 3 = 13*0 + 3, when 3 is divided by 13, we get a remainder of 3.
=> Since 9 = 13*0 + 9, when 9 is divided by 13, we get a remainder of 9.
=> Since 27 = 13*2 + 1, when 27 is divided by 13, we get a remainder of 1.
=> Since 81 = 13*6 + 3, when 81 is divided by 13, we get a remainder of 3.

So, the pattern of the remainders is 3, 9, 1, 3, 9, 1…

'm' is the remainder when  3^{2020}  is divided by 13. Since 2020 = 673*3 + 1, we get that the remainder when  3^{2020}  is divided by 13 is equal to the remainder when 31 is divided by 13. Therefore, we get a remainder of 3 when  3^{2020}  is divided by 13 and m = 3.

'n' is the remainder when  3^{2021}  is divided by 13. Since 2021 = 673*3 + 2, we get that the remainder when  3^{2021}  is divided by 13 is equal to the remainder when 32 is divided by 13. Therefore, we get a remainder of 9 when  3^{2021}  is divided by 13 and n = 9.

Therefore,   m + n = 3 + 9 = 12   and D is the correct answer.

Answer D

mehro023 · Answer

Hi, I do get the method, I am however confused with one small detail You say remainder when 3^2020 is divided by 13 is the same as remainder when 31 is divided by 13, isnt that 5 ? Similarly for 3^2021 too, doesnt 32/13 give remainder of 6 ? Am I missing something ? Your insight would be helpful

-K

VERBAL1 · Answer

My bad, I didn't interpret your solution properly.

Good luck.

A_Nishith · Answer

Q) 'm' is the remainder when 3^2020 is divided by 13 and 'n' is the remainder when 3^2021 is divided by 13.
 What is the value of m + n?
 A.6 B.8 C.10 D.12 E. 14

Given,
'm' is the remainder when 3^2020 is divided by 13 =>  3^2020 = 13x + m   (where let 'x' be Quotient)
'n' is the remainder when 3^2021 is divided by 13 =>  3^2021 = 13y + n   (where let 'y' be Quotient)

Now by comparing, We can get
3^2021 =   * 3
=> 13y + n =   * 3
=>  13y + n = 13*3x + 3m 
Comparing both sides, we can get,
y=3x &  n=3m

We need to answer of n+m=?
=>  n+m = 3m+m = 4m => as m is an integer so the answer must be the multiple of 4 ; so 8 & 12 are only possible.

Now, if n+m=8 then  m= 2   or
& if n+m=12 then  m=3

Lets find the value of 'm' :
 Remainder finding method   b^r ]

'm' is the remainder when 3^2020 is divided by 13

m = remainder of {(3^r) /13} 
remainder of 2020/4 is ZERO , so taking  r=4  {if remainder remainder is something other than 0 then 'r' would have been that value}
=> 3^4 = 81 when divided by 13 gives remainder as 3.
Therefore ,  m=3 
So,  n+m = 3m+ m = 4m = 4*3 =  12

Answer: D

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82823 GMAT-Club-Forum-np54ezi1.png

PSKhore · Answer

The easiest way to do this, and which I apply to every reminder problem:
Which power to 3 is closest to the divisor or multiple of divisor.
Here, 3^3 = 27
13X2 = 26

Hence, cycle of 3, so divide 2020 and 2021 by 3 and see how many 3's are left and just add them
In this way (1)*3 + (1)*9 will be left which will give you 12

'm' is the remainder when 3^2020 is divided by 13 and 'n' is the remai

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'm' is the remainder when 3^2020 is divided by 13 and 'n' is the remai

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