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Cool goIITian

Joined:	27 Aug 2007
Post:	59

15 Sep 2007 17:49:33 IST

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593

remainder

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

The number of positive integers which divide 2004 to leave a remainder of 24 is

A) 36

B) 20

C) 22

D) 21

E) 34

Comments (4)

goutam chalasani

Cool goIITian

Joined: 30 Aug 2007

Posts: 88

15 Sep 2007 19:41:44 IST

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ans is 20

if the integer dividing 2004 giving a remainder 24
it means it is a factor of 1980
and also it must be greater than 24

Priyesh

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Joined: 18 Feb 2007

Posts: 1042

15 Sep 2007 19:44:44 IST

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The numbers which leave remainder as 24 on dividing 2004 must be factors of (2004-24)=1980.

These factors should also be greater than 24,
otherwise they will leave a remainder <24.

The required numbers are 20.
These are given below:

30,36,44,45,55,66,90,99,110,132,165,180,198,220,330,396,495,660,990,1980.

Avinash Sharma

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Joined: 9 Mar 2007

Posts: 259

15 Sep 2007 23:34:58 IST

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Hi friends,

Your method of solving the problem is right:

M*N + R = 2004 and R=24 so

M*N = 2004-24 = 1980 now the facors of 1980 are 2,2,3,3,5,11 now take single, two at a time, three at a time ....... and all six in a go. and you will find the values of M easily.

The answer of this question is (c) 22.

Priyesh has left two numbers viz. 33 and 60. rest all (30,36,44,45,55,66,90,99,110,132,165,180,198,220,330,396,495,660,990,1980) are correct.

Priyesh

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Joined: 18 Feb 2007

Posts: 1042

16 Sep 2007 11:44:37 IST

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oh yes i left two nos

avinash sir have explained it perfectly

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