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

2^20 = 2^10 x 2^10

And 2^10 = 1024

4*4= 16, so the unit digit of this multiplication is 6

The remainder is 6

Like the binomial expression.. good job!

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

\(\frac{{2^1}}{5}\) = 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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[quote="broilerc"]What is the remainder when 2^20 is divided by 10 ?

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

2^20 can be written as (2^4)^5

16^5=16*16*16*16*16

any no having unit digit 6 multiplied n no. of times will allways give you unit digit 6. so the unit digit of this no will be 6.
any no. That is divided by 10 will allways give you unit digit as a reminder.

answer is D

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

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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units digit cycle for 2^1 through 2^4=2,4,8,6
because 20 divides by 4,
units digit for 2^20=6
6/10 gives a remainder of 6

For any given number, when divide by 10, the last digit is basically the remainder of the operation.
We have to find the last digit of 2^20.

Cyclicity of 2 is also 4:

2^1-> 2
2^2-> 4
2^3-> 8
2^4->6
2^5->2...

Here 2^20 is basically (2^5)^4 so last digit 6.

I rewrote 2^20 as 2^10^2
2^10 is 1024

4x4 = 16
and 16x16 = 256

thus when 256 is divided by 10, the remainder will be 6. is this logic appropriate?

Yes correct, just rewriting the highlighted part for you: (2^10)^2.

2^10^2 might mean 2^100. Cheers.

amegupte0410

So if the number is completely divisible, then we always take the last power?
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Asked: What is the remainder when 2^20 is divided by 10 ?

2^20mod10 = 2^{5*4}mod10 = 32^4mod10 = 2^4mod10 = 16mod10 = 6

IMO D
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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.