What is the remainder when 2^20 is divided by 10 ? : Problem Solving (PS)

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06 May 2016, 08:07

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broilerc

What is the remainder when 2^20 is divided by 10 ?

A. 0
B. 2
C. 4
D. 6
E. 8

\(\frac{{2^{20}}}{10}\) = \(\frac{{2^{20}}}{2*5}\) = \(\frac{{2^{19}}}{5}\)

\(\frac{{2^1}}{5}\) = 2

\(\frac{{2^2}}{5} = 4\)

\(\frac{{2^3}}{5} = 3\)

\({2^{19}}\) = \({2^{3*6}}\) x \(2^1\)

\(\frac{{2^{3*6}}}{5}\) = Remainder 3

\(\frac{2^1}{5}\) = Remainder 2

So, Result will be 3*2 = 6 , Answer will be (D)

PS: Its better to avoid this approach during actual GMAT exam, just posting an alternate approach for educational purpose.

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Abhishek009

broilerc

What is the remainder when 2^20 is divided by 10 ?

A. 0
B. 2
C. 4
D. 6
E. 8

\(\frac{{2^{20}}}{10}\) = \(\frac{{2^{20}}}{2*5}\) = \(\frac{{2^{19}}}{5}\)

\(\frac{{2^1}}{5}\) = 2

\(\frac{{2^2}}{5} = 4\)

\(\frac{{2^3}}{5} = 3\)

\({2^{19}}\) = \({2^{3*6}}\) x \(2^1\)

\(\frac{{2^{3*6}}}{5}\) = Remainder 3

\(\frac{2^1}{5}\) = Remainder 2

So, Result will be 3*2 = 6 , Answer will be (D)

PS: Its better to avoid this approach during actual GMAT exam, just posting an alternate approach for educational purpose.

How come you can't just take 2^19/5 and say that it is remainder 3? Isn't 2^20/10 = 2^19/5? Yet you get a remainder of 6 for the first one and a 3 for the second. I'm confused how you knew to multiply the remainder of 2^18/5 and 2/5.

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Abhishek009

broilerc

What is the remainder when 2^20 is divided by 10 ?

A. 0
B. 2
C. 4
D. 6
E. 8

\(\frac{{2^{20}}}{10}\) = \(\frac{{2^{20}}}{2*5}\) = \(\frac{{2^{19}}}{5}\)

\(\frac{{2^1}}{5}\) = 2

\(\frac{{2^2}}{5} = 4\)

\(\frac{{2^3}}{5} = 3\)

\({2^{19}}\) = \({2^{3*6}}\) x \(2^1\)

\(\frac{{2^{3*6}}}{5}\) = Remainder 3

\(\frac{2^1}{5}\) = Remainder 2

So, Result will be 3*2 = 6 , Answer will be (D)

PS: Its better to avoid this approach during actual GMAT exam, just posting an alternate approach for educational purpose.

How come you can't just take 2^19/5 and say that it is remainder 3? Isn't 2^20/10 = 2^19/5? Yet you get a remainder of 6 for the first one and a 3 for the second. I'm confused how you knew to multiply the remainder of 2^18/5 and 2/5.

Simplification changes the remainder. Look at this:

25/10 - Remainder 5
But
5/2 - Remainder 1

Dividend = Quotient * Divisor + Remainder

Like in the example above, when dividend and divisor are divided by 5, the Remainder gets divided by 5 too. So to get the actual Remainder, you need to multiply the Remainder by 5 again.

Hence when you use 2^19/5 (after dividing both Dividend and Divisor by 2) and get the remainder 3, you need to multiply it by 2 back to get the remainder 6.

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01 Jan 2017, 14:15

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Easy explanation:

\(2^{20}=4^{10}=16^5\)

All powers of 6 end with a 6 in the units digit, so \(16^5\) must also end with a 6. Thus, when divided by 10, the remainder must be 6.

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12 Jan 2017, 17:07

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mcelroytutoring

Easy explanation:

\(2^{20}=4^{10}=16^5\)

All powers of 6 end with a 6 in the units digit, so \(16^5\) must also end with a 6. Thus, when divided by 10, the remainder must be 6.

mcelroytutoring Could you briefly explain this? I am not understand about this.

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mcelroytutoring

Easy explanation:

\(2^{20}=4^{10}=16^5\)

All powers of 6 end with a 6 in the units digit, so \(16^5\) must also end with a 6. Thus, when divided by 10, the remainder must be 6.

mcelroytutoring Could you briefly explain this? I am not understand about this.

Sure! When you multiply 20 2s together, you get 10 4s because each pair of 2s makes a 4. (2 x 2 = 4) When you multiply 10 4s together, you get 5 16s because every pair of 4s makes a 16 (4 x 4 = 16).

Every power of 6 ends with a 6 because \((6)(6) =36\) and \((6)(6)(6) = (36)(6) = 216\) and \((6)(6)(6)(6) = 1296\), etc. And the remainder when you divide by 10 will always be equal to 6, because multiples of 10 always end in 0. \(36/10 = 3 R 6, 216/10 = 21 R 6, 1296/10 = 129 R 6\), etc.

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Remainder of ANY number upon division with 10 will be the UNIT'S DIGIT of the number (dividend).

e.g. 22 divided by 10 will give 2 as the remainder which is the unit's digit.

Back to the question:
2^20 divided by 10

Cyclicity rules come in handy here:
Cyclicity of 2 is 4 that means the unit's digits start repeating after every 4th number.

2^1=2
2^2=4
2^3=8
2^4=16 Unit's digit=6
2^5=32 Unit's digit=2 (Started repeating)

Here the power of 2 is 20.
As the cyclicity of 2 is 4, we will divide 20 by 4 which gives us a remainder of 0.
Thus the unit's digit (here) would be 6.

Therefore the remainder upon division with 10 is 6.

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broilerc

What is the remainder when 2^20 is divided by 10 ?

A. 0
B. 2
C. 4
D. 6
E. 8

We need to determine the remainder when 2^20 is divided by 10. To do so, recall that any number divided by 10 will produce the same remainder as the units digit of that number. Thus, let’s determine the units digit of 2^20. The pattern of units digits of 2 when raised to a positive integer exponent is:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 6

2^5 = 2

We see that the pattern is 2-4-8-6. Furthermore, 2^4k, in which k is a positive integer, will always have a units digit of 6.

Thus, the units digit of 2^20 is 6. Dividing 6 by 10 yields a remainder of 6; thus, dividing 2^20 by 10 also yields a remainder of 6.

Answer: D

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broilerc

What is the remainder when 2^20 is divided by 10 ?

A. 0
B. 2
C. 4
D. 6
E. 8

Hi,

TWO ways-

1) Cyclic pattern of units digit-
Remainder when div by 10 is nothing BUT units digit
\(2^1 = 2.........\\
2^2 = 4..........\\
2^3 = 8..........\\
2^4 = 16.. or... 6........\)
and this carries on in same pattern.... 2, 4, 8, 6, 2, 4, 8, 6... so 20 is div by 4..
so 2^20 will have UNITS digit same as 4th power..
ans 6

2) binomial expansion
\(2^{20} = (2^5)^4 = 32^4 = (30+2)^4\)..
Now the above expression will have all other terms div by 10 except 2^4...
\(2^4 = 16\)..
and 16 div by 10 gives a remainder of 6..

D

Hi! Are there any circumstances that we can't use cyclicity? I love using the binomial expansion, but sometimes can't figure them out.

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Hi! Are there any circumstances that we can't use cyclicity? I love using the binomial expansion, but sometimes can't figure them out.

Cyclicity is the best method, as when we divide by 10, we want the last digit.

In this case, we can break up 2^20 as 2^5 * 2^5 * 2^5 * 2^5 = 32 * 32 * 32 * 32.

The last digit is the multiplication of all the units place = 2 * 2 * 2* 2 = 16. Therefor the last digit is 6.

Hope This helps

Arun Kumar

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This is a GMAT prep exam pack 2 question. I got it on mock #6. Bunuel please add the tag.

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This is a GMAT prep exam pack 2 question. I got it on mock #6. Bunuel please add the tag.

_____________________
Added the tag. Thank you.

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Updated on: 21 Sep 2024, 22:53

Originally posted by BrushMyQuant on 07 Sep 2022, 12:04.
Last edited by BrushMyQuant on 21 Sep 2024, 22:53, edited 2 times in total.

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What is the remainder of \(2^{20}\) when divided by 10

Theory: Remainder of a number by 10 is same as the unit's digit of the number

(Watch this Video to Learn How to find Remainders of Numbers by 10)

Using Above theory Remainder of \(2^{20}\) by 10 = unit's digit of \(2^{20}\)

Now to find the unit's digit of \(2^{20}\), we need to find the pattern / cycle of unit's digit of power of 2 and then generalizing it.

Unit's digit of \(2^1\) = 2
Unit's digit of \(2^2\) = 4
Unit's digit of \(2^3\) = 8
Unit's digit of \(2^4\) = 6
Unit's digit of \(2^5\) = 2

So, unit's digit of power of 2 repeats after every \(4^{th}\) number.
=> We need to divided 20 by 4 and check what is the remainder
=> 20 divided by 4 gives 0 remainder

=> \(2^{20}\) will have the same unit's digit as \(2^4\) = 6

So, Answer will be D
Hope it helps!

MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

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Hello. Good Day to you all. Please, I am finding the quant very difficult(in general). I am very good on verbal. But quant is a huge problem for me, please what do you all suggest I do? I have one month left to take the GMAT.

I will be waiting your response. Thanks.

Posted from my mobile device

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I did this
(2^4)^5/10 = (20-4)^5/10 =

4 patter is 4,6,4,6,4
So shouldn't the remainder be 4?

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Updated on: 21 Sep 2024, 22:57

Originally posted by BrushMyQuant on 21 Sep 2024, 22:53.
Last edited by BrushMyQuant on 21 Sep 2024, 22:57, edited 1 time in total.

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hellohnfjskfbkse

I did this
(2^4)^5/10 = (20-4)^5/10 =

4 patter is 4,6,4,6,4
So shouldn't the remainder be 4?

If you were trying to convert the problem from power of 2 to power to 4 then following is the step

\(2^{20}\) = \(2^{(2*10)}\) = \(4^{10}\) [ You converted this to \( (2^4)^5 \) = \((16)^5\) = \((4)^{10}\) ≠ \((4)^5\) ]

Unit's digit of \(4^1\) = 4
Unit's digit of \(4^2\) = 6
Unit's digit of \(4^3\) = 4
Unit's digit of \(4^4\) = 6

=> Units' digit = 4 when power is odd
=> Units' digit = 6 when power is even

=> Units' digit of \(4^{10}\) = 6

So, Answer will be D
Hope it helps!

MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

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Hello. Good Day to you all. Please, I am finding the quant very difficult(in general). I am very good on verbal. But quant is a huge problem for me, please what do you all suggest I do? I have one month left to take the GMAT.

I will be waiting your response. Thanks.

Posted from my mobile device

I wrote this post sometime back, hope you find it useful

https://gmatclub.com/forum/how-to-impro ... 37975.html

Additionally I post frequent videos on YouTube, you can follow the same here BrushMyQuant

Hope it helps!

What is the remainder when 2^20 is divided by 10 ?

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