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

Question

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

JeffTargetTestPrep · Accepted Answer

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

nipunjain14 · Answer

999^500 will have units digit 1 thus when divided by 5 will leave reminder 1.
Ans: B

ronny123 · Answer

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

Abhishek009 · Answer

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)

ion · Answer

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

Thank you in advance for answer

KrishnakumarKA1 · Answer

999^500 can be written as (1000-1)^500. Remainder when divided by 5 will be (-1)^500 or 1. Option B

nymfan14 · Answer

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?

gracie · Answer

units digit cycle for powers of 9:
9^odd power=9
9^even power=1
1/5 gives a remainder of 1
B

generis · Answer

@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)

Feb2024 · 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?

Bunuel · Answer

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.

BrushMyQuant · Answer

We know to find what is the remainder when   999^{500}   is divided by 5

Lets solve the problem using two methods:

Method 1: Binomial Theorem

We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(5) and a big number, other part is a small number.

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

Watch this video to   MASTER BINOMIAL Theorem

Now, when we expand this expression then all the terms except the last term will be a multiple of 5. 
=> All terms except the last term will give 0 as remainder then divided by 5
=> Problem is reduced to what is the remainder when the last term is divided by 5
=> What is the remainder when  500C500 * 1000^0 * (-1)^ {500}  is divided by 5 = Remainder of 1 by 5 = 1

Method 2: Units' Digit Cycle

Theory: Remainder of a number by 5 is same as the Remainder of the unit's digit of the number by 5

( Watch this Video  to Learn  How to find Remainders of Numbers by 5 )

Using Above theory , Let's find the unit's digit of  999^{500}  first.
Units' digit of  999^{500}  will be same as units' digit of  9^{500}

We can do this by finding the pattern / cycle of unit's digit of power of 9 and then generalizing it.

Unit's digit of  9^1  = 9
Unit's digit of  9^2  = 1
Unit's digit of  9^3  = 9
Unit's digit of  9^4  = 1

So, unit's digit of power of 9 repeats after every  2^{nd}  number.
=> If power is odd then units' digit is 9
=> If power is even then units' digit is 1

=> Unit's digits of  9^{500}  = 1
=> Remainder of  999^{500}  by 5 = 1

So,  Answer will be B 
Hope it helps!

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

https://www.youtube.com/watch?v=h1HTEhmUb1w

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards