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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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New post 29 Apr 2016, 16:06
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A
B
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D
E

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72% (00:54) correct 28% (01:00) wrong based on 267 sessions

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

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New post 30 Apr 2016, 11:45
999^500 will have units digit 1 thus when divided by 5 will leave reminder 1.
Ans: B
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Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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New post 30 Apr 2016, 16:57
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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New post 01 May 2016, 02:27
2
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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New post 02 Mar 2017, 10:56
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?

Thank you in advance for answer
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Re: When 999^500 is divided by 5, the remainder is  [#permalink]

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New post 07 Mar 2017, 17:41
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.

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

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New post 08 Mar 2017, 03:42
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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New post 18 Apr 2018, 09:39
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.

Answer: B

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

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New post 19 Apr 2018, 11:50
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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When 999^500 is divided by 5, the remainder is  [#permalink]

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New post 20 Apr 2018, 20:28
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.

Answer: B

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?

@nymfan , glad you posted.

If you want to tag a specific person to answer a question, type the "@" with their username immediately after (no space between "@" username) and then, leave one space after those two things (do not put a comma right after the name , for example).

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You made a small mistake.

\(9^5\) 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, 20:28
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