IIT JEE , AIEEE Forum - Mathematics , Algebra

prathima

if m&n r natural numbers such that

m/n=1/3+1/5+1/17+1/19+1/23+1/1979+1/1983+1/1985+1/1997+1/1999

will2001 or 2002 divides m?????

puneet1622

Yaar check ur question pls ... u can always say that 2001 and 2002 will divide m .

If 2001 does not divide m then multiply both m and n by 2001. Now the m/n ratio remains same but m is divisible by 2001.

So we are done ...

But i think the question demands something else .. check it out .

cheers

gorakavipraveen

edison

I agree upon the logic suggested by Puneet1622, please mention whether m/n is the simplest ratio or any other ratio.

polymath

m/n= 1/3 + 1/1999 + 1/5 + 1/1997 + 1/17 + 1/1985 + 1/19 + 1/1983 + 1/23 + 1/1979

m/n = 1999+3/3*1999 + 1997+5/5*1997 + 1985+17/17*1985 + 1983+19/19*1983 + 1979+23/23*1979

taking 2002 common

m/n = 2002 (1/3*1999 + 1/5*1997 + 1/17*1985 + 1/ 19*1983 + 1/ 23*1979)

let 1/3*1999 + 1/5*1997 + 1/17*1985 + 1/ 19*1983 + 1/ 23*1979 = k

m/n = 2002*k

m/n is divisible by 2002. This is how we can approach this problem. But k may

or may not be an integer so we cant say surely wether it is divisible by 2002 or

not.Since you have given two options i ll assume k is a number. so 2002 is right

choice.

edison

Well done polymath keep it up. The way u cracked the problem is fantastic and appreciated

gorakavipraveen

hey polymath
whats your opinion on your build up constant k. I did the same but with the same( following) thought I have discharged my idea 2 hours ago.
Confirm me what exactly m/n gives. If it is a fraction, then no mind ,all m and n (I MEAN ALL CONSTANTS ARE DIVISIBLE BY ALL NUMBERS .HERE CONSTANT I MEAN IS IF WE CONSIDER M OR N AS CONSTANTS FOR A PERTICULAR CASE.)can be divided by what not every constant. For the given case its idle to condsider as m/2001 or m/2002 NOT A FRACTION.I THINK THIS IS THE REQUIRED CONDITION FOR A VARIABLE TO BE PERFECTLY DIVIDED BY SOME OTHER. So in accordanse to PUNEET, some data is missed. If your proof is climed correct, iI guess 1 can be dividedby 1,03,89,431 and what not every element on number line.so probably k according to me can not be a fraction as against your proof. I am sorry to say this but I am trying to explore more through this healthy debate.

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