playing with p,q,r

Number Theory Level pending

If a number is in a form of - 100p+10q+r + 100r+10q+p Then by which variable it would definately be divisible? And why? Given p,q,r are consecutive numbers.

q all p r

Guneet Singh Oberai by Guneet Singh Oberai , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

David Nichols
Jul 21, 2015

First of all I wanted an expression with just 1 variable.

Since they are consecutive we can say q=p+1 and r=p+2.

Substituting and expanding gives us 222p +222 or 222(p+1).

We know q=p+1 so q is definitely a factor and we can't say for certain that p or r are factors.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Because, as p,q,r are consecutive number the equation made would be like this - p=q-1,q=q,r=q+1. therefore after solution q would be coming as one of the factor

0 Helpful 0 Interesting 0 Brilliant 0 Confused

How do we know that there exists no other factor from the options?

Sharky Kesa - 5 years, 10 months ago

Log in to reply

Take a eg. - 789+987= 1776 here q = 8 and 1776 is perfectly divisible by 8 ie. 1776/8=222 but not perfectly divisible by 7,9 therefore in any case q would be a factor

Guneet Singh Oberai - 5 years, 10 months ago

Log in to reply

Giving a specific example does not prove for all.

Sharky Kesa - 5 years, 10 months ago

Log in to reply

@Sharky Kesa – As previously explained p=q-1, q=q , r=q+1 therefore . Any real number n = 100q-100+10q+q+1+ 100q+100+10q+q-1= 200q+20q+2q = q ( 222 ) . Thus q is only a factor of n not any other variable.

Guneet Singh Oberai - 5 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...