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# When 999^500 is divided by 5, the remainder is

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When 999^500 is divided by 5, the remainder is  [#permalink]

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29 Apr 2016, 15:06
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Difficulty:

15% (low)

Question Stats:

70% (00:54) correct 30% (00:57) wrong based on 271 sessions

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When 999^500 is divided by 5, the remainder is

A. 0
B. 1
C. 2
D. 3
E. 4

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Joined: 07 May 2015
Posts: 175
Location: India
GMAT 1: 660 Q48 V31
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Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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30 Apr 2016, 10:45
999^500 will have units digit 1 thus when divided by 5 will leave reminder 1.
Ans: B
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Joined: 12 Jun 2015
Posts: 79
Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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30 Apr 2016, 15:57
999^500 is to be divided by 5

We can reduce the number to 4^500 (as the remainder of 999/5 is 4 and we can get the same remainder for 4^500 and 999^500 regardless of which is divided 5 )

Now, let's notice the pattern of remainders of powers of 4 when divided by 5

R{(4^1)/5} = 4
R{(4^2)/5} = 1
R{(4^3)/5} = 4
R{(4^4)/5} = 1
and so on

So, even powers of 4 give a remainder of 1
Hence, R{(4^500)/5} will be 1
R{(999^500)/5} will also be 1

Correct Option : B
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Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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01 May 2016, 01:27
3
Bunuel wrote:
When 999^500 is divided by 5, the remainder is

A. 0
B. 1
C. 2
D. 3
E. 4

999^500 = $$(1000 - 1)^{500}$$

1000/5 will leave no remainder

-1^{Even} will be 1

1/5 will leave remainder 1

So, answer will be 1 , option (B)
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Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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02 Mar 2017, 09:56
Abhishek009 wrote:
Bunuel wrote:
When 999^500 is divided by 5, the remainder is

A. 0
B. 1
C. 2
D. 3
E. 4

999^500 = $$(1000 - 1)^{500}$$

1000/5 will leave no remainder

-1^{Even} will be 1

1/5 will leave remainder 1

So, answer will be 1 , option (B)

But in case -1 had an odd power, in that case it would be 4?

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Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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07 Mar 2017, 16:41
1
Bunuel wrote:
When 999^500 is divided by 5, the remainder is

A. 0
B. 1
C. 2
D. 3
E. 4

We may recall that when dividing a number by 5, the remainder will be the same as when we divide the units digit of that number by 5. Thus, to determine the remainder when 999^500 is divided by 5, we need to get the units digit of 999^500, or simply 9^500.

Let’s start by evaluating the pattern of the units digits of 9^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 9. When writing out the pattern, notice that we are ONLY concerned with the units digit of 9 raised to each power.

9^1 = 9

9^2 = 1

9^3 = 9

9^4 = 1

The pattern of any base of 9 repeats every 2 exponents. The pattern is 9–1. In this pattern, all positive exponents that are even will produce a 1 as its units digit. Thus:

9^500 has a units digit of 1, and since 1/5 has a reminder of 1, the remainder when 999^500 is divided by 5 is 1.

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Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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08 Mar 2017, 02:42
999^500 can be written as (1000-1)^500. Remainder when divided by 5 will be (-1)^500 or 1. Option B
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Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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18 Apr 2018, 08:39
JeffTargetTestPrep wrote:
Bunuel wrote:
When 999^500 is divided by 5, the remainder is

A. 0
B. 1
C. 2
D. 3
E. 4

We may recall that when dividing a number by 5, the remainder will be the same as when we divide the units digit of that number by 5. Thus, to determine the remainder when 999^500 is divided by 5, we need to get the units digit of 999^500, or simply 9^500.

Let’s start by evaluating the pattern of the units digits of 9^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 9. When writing out the pattern, notice that we are ONLY concerned with the units digit of 9 raised to each power.

9^1 = 9

9^2 = 1

9^3 = 9

9^4 = 1

The pattern of any base of 9 repeats every 2 exponents. The pattern is 9–1. In this pattern, all positive exponents that are even will produce a 1 as its units digit. Thus:

9^500 has a units digit of 1, and since 1/5 has a reminder of 1, the remainder when 999^500 is divided by 5 is 1.

I saw the 1/9 pattern you outlined above but 9^5 has a unit digit of 9. 9/5 = r4 so I chose E. What am I missing?
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Joined: 07 Dec 2014
Posts: 1151
Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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19 Apr 2018, 10:50
Bunuel wrote:
When 999^500 is divided by 5, the remainder is

A. 0
B. 1
C. 2
D. 3
E. 4

units digit cycle for powers of 9:
9^odd power=9
9^even power=1
1/5 gives a remainder of 1
B
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Joined: 22 May 2016
Posts: 2344
When 999^500 is divided by 5, the remainder is  [#permalink]

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20 Apr 2018, 19:28
JeffTargetTestPrep wrote:
Bunuel wrote:
When 999^500 is divided by 5, the remainder is

A. 0
B. 1
C. 2
D. 3
E. 4

We may recall that when dividing a number by 5, the remainder will be the same as when we divide the units digit of that number by 5. Thus, to determine the remainder when 999^500 is divided by 5, we need to get the units digit of 999^500, or simply 9^500.

Let’s start by evaluating the pattern of the units digits of 9^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 9. When writing out the pattern, notice that we are ONLY concerned with the units digit of 9 raised to each power.

9^1 = 9

9^2 = 1

9^3 = 9

9^4 = 1

The pattern of any base of 9 repeats every 2 exponents. The pattern is 9–1. In this pattern, all positive exponents that are even will produce a 1 as its units digit. Thus:

9^500 has a units digit of 1, and since 1/5 has a reminder of 1, the remainder when 999^500 is divided by 5 is 1.

nymfan14 wrote:
I saw the 1/9 pattern you outlined above but 9^5 has a unit digit of 9. 9/5 = r4 so I chose E. What am I missing?

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$$9^5$$ has a units digit of 9, as does every odd power of 9. And true, a units digit of 9 will leave a remainder of 4.

But $$9^{500}$$ is an even power. Every even power of 9 has a units digit of 1 (and, if divided by 5, will leave a remainder of 1). Easy mistake.

I hope that helps.

For more on cyclicity, see Bunuel Last Digit of a Power (scroll down)
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When 999^500 is divided by 5, the remainder is &nbs [#permalink] 20 Apr 2018, 19:28
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