Arithmetic sequence

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danjuma: Senior | Next Rank: 100 Posts; Posts: 89; Joined: Thu Aug 26, 2010 9:29 am; Thanked: 4 times

by danjuma » Thu Dec 09, 2010 12:03 pm

What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?

1.11

2.17

3.19

4.23

5.35

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Source: Beat The GMAT — Problem Solving | Thu Dec 09, 2010 12:03 pm

N:Dure: Master | Next Rank: 500 Posts; Posts: 181; Joined: Sat Oct 16, 2010 1:57 pm; Thanked: 4 times

by N:Dure » Thu Dec 09, 2010 1:04 pm

I believe the answer is 3) 19

19 /4 = 4 + remainder 3

17 /3 = 5 + 2

15/ 2 = 7 + 1

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Thu Dec 09, 2010 1:08 pm

danjuma wrote:What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?

1.11

2.17

3.19

4.23

5.35

There can be Many value for the third term in the series. Because we don't know what is the first term of the sequence.

it can be 11, 23 , 35,

It can be -11, -23, -35, and with the difference of 12 there can be n number of values.

I know as per the options 35 seems to be first choice. But What if negative values.

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Thu Dec 09, 2010 8:11 pm

danjuma wrote:What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?

1.11

2.17

3.19

4.23

5.35

Please post the OA and the explanation if you have with you?

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

by diebeatsthegmat » Fri Dec 10, 2010 5:31 am

danjuma wrote:What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?

1.11

2.17

3.19

4.23

5.35

i think the answer is 11, because 11 : 2= 5 with remainder 1
11:3 =3 with the remainder 2
and 11:4 =2 with remainder 3
in case the number is the same
in case the number are different from the sequence, 19 or 23 also be the answers
there's something not right with this PS
experts please!

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kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members; Location: madrid; Thanked: 171 times; GMAT Score:790

by kevincanspain » Fri Dec 10, 2010 5:36 am

Instead of 'numbers', the question should read ' positive integers'
Also, it should be made clear that this is an increasing sequence.

Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid

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