-1 divisibilty problem

Number Theory Level 2

The Positive Integers A and B Which \textit{The Positive Integers A and B Which} A B \frac{A}{B} , A − 1 B − 1 \frac{A-1}{B-1} , A − 2 B − 2 \frac{A-2}{B-2} give an integer \textit{give an integer}

A ≠ B ≠ ( 1 , 2 , 3 ) {A}\neq{B} \neq{(1,2,3) }

For example : 9 3 \frac{9}{3} gives integer and \textit{gives integer and} 9 − 1 3 − 1 \frac{9-1}{3-1} also gives integer \textit{also gives integer} and this only 2 fractions \textit{and this only 2 fractions}

  • The Question Is \textbf{The Question Is} What is the smallest 2 integers which \textbf{What is the smallest 2 integers which} A B \frac{A}{B} , A − 1 B − 1 \frac{A-1}{B-1} , A − 2 B − 2 \frac{A-2}{B-2} (3 Fractions) gives integer? \textbf{gives integer?}

A ≠ B ≠ ( 1 , 2 , 3 ) {A}\neq{B} \neq{(1,2,3) }

  • The awnser is the Sum of A,B \textit{The awnser is the Sum of {A,B}}


The answer is 20.

Ahmed Pro by Ahmed Pro , 18, Egypt

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.

1 solution

Ahmed Pro
Jul 27, 2020

1 \boxed{1} X n − 1 X m − 1 \dfrac{X^{n}-1}{X^{m}-1} = = Integer, When n m \dfrac{n}{m} = = Integer

Also X n X m \dfrac{X^n} {X^m} = = X n − m X^{n-m} = = an integer

  • So 2 X n − 2 2 x m − 2 \dfrac{2^{X^{n}}-2}{2^{x^{m}}-2} = = 2 × ( 2 X n − 1 − 1 ) 2 × ( 2 x m − 1 − 1 ) \dfrac{2×(2^{X^{n}-1}-1)}{2×(2^{x^{m}-1}-1)} = = 2 X n − 1 − 1 2 x m − 1 − 1 \dfrac{2^{X^{n}-1}-1}{2^{x^{m}-1}-1} ⟶ \longrightarrow Because it is same to equation 1 \boxed{1}

  • So we can apply the same rule which 2 X n − 2 2 x m − 2 \dfrac{2^{X^{n}}-2}{2^{x^{m}}-2} = = 2 X n − 1 − 1 2 x m − 1 − 1 \dfrac{2^{X^{n}-1}-1}{2^{x^{m}-1}-1} = = An integer When X n − 1 X m − 1 \dfrac{X^n-1}{X^m-1} = = Integer

Now we know that { 2 X n − 2 2 x m − 2 \dfrac{2^{X^{n}}-2}{2^{x^{m}}-2} , 2 X n − 1 2 x m − 1 \dfrac{2^{X^{n}}-1}{2^{x^{m}}-1} , 2 X n 2 x m \dfrac{2^{X^{n}}}{2^{x^{m}}} } = = integer when { x n x m \dfrac{x^n}{x^m} , x n − 1 x m − 1 \dfrac{x^n-1}{x^m-1} } = = integer

  • { x n x m \dfrac{x^n}{x^m} , x n − 1 x m − 1 \dfrac{x^n-1}{x^m-1} } = = integer when n m \dfrac{n} {m} = = integer

⟹ \implies So { 2 X n − 2 2 x m − 2 \dfrac{2^{X^{n}}-2}{2^{x^{m}}-2} , 2 X n − 1 2 x m − 1 \dfrac{2^{X^{n}}-1}{2^{x^{m}}-1} , 2 X n 2 x m \dfrac{2^{X^{n}}}{2^{x^{m}}} } = = integer, when n m \dfrac{n} {m} = = integer

  • So \textbf{So} { 2 X n , 2 X m 2^{X^n} , 2^{X^m} } =(A,B) When n m \dfrac{n} {m} =integer

Let X be the smallest integer we can choose

When X=1, The 2 numbers will be = 2 and A ≠ \neq B

So X = 2 \boxed{X=2}

And n ≠ \neq m because A ≠ \neq B So the smallest possible 2 integers which not equal and when dividing them they gives an integer is (2,1)

So ( n , m ) = ( 2 , 1 ) \boxed{ (n, m) =(2, 1)}

And by putting X, n, m in their places

(A, B) = ( 2 X n , 2 X m 2^{X^n} , 2^{X^m} ) = ( 2 2 2 , 2 2 1 2^{2^2} , 2^{2^1} ) = (16,4)

( A , B ) = ( 4 , 16 ) \boxed{(A, B) =(4, 16)}

  • And by sum A and B = 20
1 Helpful 1 Interesting 1 Brilliant 1 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...