BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Number properties

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Sep 20, 2015 4:30 pm
sud21 wrote:What is the remainder when 10^ 49 +2 is divided by 11 ?

1

2

3

5

7
Unless I'm missing a completely different solution, this question relies on the rule for determining whether a number is divisible by 11 (https://www.math.hmc.edu/funfacts/ffiles/10013.5.shtml). I've never seen an official question rely on this rule, so in my opinion, it's out of scope for the GMAT.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Master | Next Rank: 500 Posts
Posts: 344
Joined: Sat Nov 12, 2011 3:21 am
Thanked: 1 times
Followed by:2 members

by sud21 » Sun Sep 20, 2015 7:22 pm
It should be read as What is the remainder when {10^49} +2 is divided by 11 ?

Can it be solved?
Top
Master | Next Rank: 500 Posts
Posts: 126
Joined: Sat Jun 07, 2014 5:26 am
Thanked: 3 times

by sandipgumtya » Mon Sep 21, 2015 6:08 am
I think the exponent rule will have a role here.
10^1+2-Remainder is 1
10^2+2-Remainder is 3
10^3+2-Remainder is 1
10^4+2-Remainder is 3
.
.
.
So on.
SO 10^49+2 will have remainder of 1.Ans-A
Top
Master | Next Rank: 500 Posts
Posts: 344
Joined: Sat Nov 12, 2011 3:21 am
Thanked: 1 times
Followed by:2 members

by sud21 » Mon Sep 21, 2015 4:14 pm
That's what I thought. You divide any 1000 or 10000, if unit digit ends in zero, then the maximum way the quotient can be is 9 and then you add 2 which becomes 3.
Top
Master | Next Rank: 500 Posts
Posts: 126
Joined: Sat Jun 07, 2014 5:26 am
Thanked: 3 times

by sandipgumtya » Mon Sep 21, 2015 7:15 pm
If u have noticed :Here every Odd power of 10 would produce remainder of 1 and Even power 3.This is the pattern.
Top
Master | Next Rank: 500 Posts
Posts: 344
Joined: Sat Nov 12, 2011 3:21 am
Thanked: 1 times
Followed by:2 members

by sud21 » Tue Sep 22, 2015 7:17 pm
The pattern is like that here.
Top
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Thu Sep 24, 2015 10:17 am
Two approaches:

1:: Divisibility by 11

A number is divisible by 11 if the difference between the alternating sums of its digits = a multiple of 11. (For instance, 1243 divides by 11, because (2 + 3) - (4 + 1) = 0. 9273 is also divisible by 11, because (9 + 7) - (2 + 3) = 11, etc.)

10�� + 2 is a 50 digit number that starts with a 1, ends with a 2, is otherwise all 0s. So its alternating sum difference is 2 - 1 = 1. This means it has a remainder of 11, and we're done.


2:: Modular arithmetic

10 has a remainder of -1 when divided by 11, so 10�� has a remainder of (-1)��, or -1, when divided by 11. Since 10�� has a remainder of -1, 10�� + 2 has a remainder of -1 + 2, or 1.
Top
Post new topic Post Reply
• Page 1 of 1