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

Also, I have to put the quotation marks around the at sign because formatting picks it up and does not show it as an at sign. Do not actually use quotation marks.

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

Bunuel Abhishek009 KarishmaB Would be thankful if you reply to the query above.

But in case -1 had an odd power, in that case would be remainder be 4?­
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When 999^500 is divided by 5, the remainder is [#permalink]
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Expert Reply
July2023 wrote:
ion wrote:
Abhishek009 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

Bunuel Abhishek009 KarishmaB Would be thankful if you reply to the query above.

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

Correct, because -1 divided by 5 gives a remainder of 4. Hence, if it were "What is the remainder when 999^501 is divided by 5," the answer would be 4.­
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