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555-605 Level|   Exponents|   Remainders|      
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Bunuel
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GMAT Club Around The World!!! What is the remainder when 3^50 is 31 Dec 2021, 11:15
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What is the remainder when \(3^{50}\) is divided by 7?

A. 1
B. 2
C. 3
D. 4
E. 5

 


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GMAT Club Around The World!!! What is the remainder when 3^50 is 02 Jan 2022, 15:00
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Bunuel
What is the remainder when \(3^{50}\) is divided by 7?

A. 1
B. 2
C. 3
D. 4
E. 5




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GMAT Club Around The World!!! What is the remainder when 3^50 is 31 Dec 2021, 11:27
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Bunuel
What is the remainder when \(3^{50}\) is divided by 7?

A. 1
B. 2
C. 3
D. 4
E. 5




Happy New Year Nepal

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3^50 can be written as (3^3)^16 * 3^2
So, (27^16 * 9) / 7
27 leaves a remainder of -1 with 7 and -1^16 = 1
1*9/7 = 2 remainder. B
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GMAT Club Around The World!!! What is the remainder when 3^50 is 31 Dec 2021, 11:37
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Euler of 7 = 6. so 3^6k mod 7 =0.
Left part is - 3^2 mod 7=9 mod 7=2 .
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GMAT Club Around The World!!! What is the remainder when 3^50 is 31 Dec 2021, 13:11
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Remainder(\(\frac{3^{50}}{7}\))

Remainder(\(\frac{3^{48}}{7}\)) * Remainder(\(\frac{3^{2}}{7}\))

Remainder(\(\frac{3^{16 * 3} }{7}\)) * Remainder(\(\frac{3^{2}}{7}\))

\(\frac{ - 1 ^{16} }{7}\) * 2

1 * 2

= 2
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GMAT Club Around The World!!! What is the remainder when 3^50 is 01 Jan 2022, 01:35
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3^50 = 9^25 ÷ 7


9^1=9÷7, leaves remainder 2 ………(1)
9^2=81÷7, leaves remainder 4 ………(2)
9^3=729÷7, leaves remainder 1 …….(3)
9^4=6561÷7, leaves remainder 2 …….(4)
9^5=59049÷7, leaves remainder 4 ……….(5)
9^6=531441÷7, leaves remainder 1 ……..(6)
9^7=4782969÷7, leaves remainder 2 ….(7)

Cyclic number : 3
Thus, 25 ÷ 3 = (reminder 1 < cyclic number)
9^25 = 9^1 when ÷ 7, reminder 2
Option B

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GMAT Club Around The World!!! What is the remainder when 3^50 is 01 Sep 2023, 10:31
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3^1 - 3 ————-
3^2 - 9 ————- 9/7 , rem = 2
3^3 - 27 ————- 27/7 , rem = 6
3^4 - 81 ————- 81/7 , rem = 4
3^5 - 243 ————- 243/7 , rem = 2
3^6- 729 ————- 729/7 , rem = 6

So cyclical pattern
So 3^50 corresponds to 3^2 will give rem 2
Op B

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GMAT Club Around The World!!! What is the remainder when 3^50 is 30 Jul 2025, 06:01
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The remainder when divided by 7 will be the following for 3^5, 3^6, 3^7 -

  • 3^5 - 243/7; R = 5
  • 3^6 - 729/7; R = 1
  • 3^7 - 2187/7; R = 3

The negative remainders will be -2, -6 & -4 respectively which is not the same as 5, 1 & 3. Hence, the cyclicity will be 6.


DiaPro
3^1 - 3 ————-
3^2 - 9 ————- 9/7 , rem = 2
3^3 - 27 ————- 27/7 , rem = 6
3^4 - 81 ————- 81/7 , rem = 4
3^5 - 243 ————- 243/7 , rem = 2
3^6- 729 ————- 729/7 , rem = 6

So cyclical pattern
So 3^50 corresponds to 3^2 will give rem 2
Op B

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GMAT Club Around The World!!! What is the remainder when 3^50 is 11 Aug 2025, 18:56
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Can someone tell if my approach is correct?

Since cyclicality of 3 is 4, I divided 40 by 4 leaving 2 as remainder. 3^2 = 9 which I divided by 7 and remainder was 2.
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GMAT Club Around The World!!! What is the remainder when 3^50 is 12 Aug 2025, 00:37
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MegB07
Can someone tell if my approach is correct?

Since cyclicality of 3 is 4, I divided 40 by 4 leaving 2 as remainder. 3^2 = 9 which I divided by 7 and remainder was 2.
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The cyclicity you refer to there determines the units digit of 3 in a positive integer power, which will not always work here. For example, if we had 3^5, then 5 gives a remainder of 1 when dividing by 4, and 3^1 divided by 7 gives a remainder of 3. However, the actual remainder when 3^5 is divided by 7 is 5.
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