# Why the following G.C.D is $1$

Why is the following G.C.D equal to $1$: $$\gcd(3^s, 2^n-3^{(j-i)}2^m),\quad s> j >i \geq 0,$$ and all variables are natural numbers.

• If $i$ ande $j$ are switched it looks right, and easy to show. – coffeemath Nov 21 '12 at 12:19
• @tlh1987 Because $i>j \implies i-j>0 \implies j-i<0 \implies 3^{j-i} \not\in \mathbb{N}$ – badp Nov 21 '12 at 12:22
• @joriki, sorry, there something wrong with the material I have, can I change the question? I am first asked here? – tlh1987 Nov 21 '12 at 12:29
• @badp sorry, I swithed the i and j. – tlh1987 Nov 21 '12 at 12:30
• @tlh1987: When you edit the question in a way that makes an existing answer or comment appear wrong, please mark the edit as such. Thanks. – joriki Nov 21 '12 at 12:55

## 2 Answers

The only prime factor of $3^s$ is 3 as $s\ge 1$

But $2^n-3^{(j-i)}2^m\equiv2^n\pmod 3$ as $3\mid 3^{j-i}$ as $j>i$

So, $2^n-3^{(j-i)}2^m\equiv2^n\equiv(-1)^n\not\equiv 0 \pmod 3$

So, $3^s,2^n-3^{(j-i)}2^m$ can not have any common prime factor, hence $(3^s,2^n-3^{(j-i)}2^m)=1$

• sorry, I switched the i and j, is this still the answer? – tlh1987 Nov 21 '12 at 12:31
• @tlh1987, please find the edited answer. – lab bhattacharjee Nov 21 '12 at 12:33
• Excuse my ignorance... what does the bar in $j>i,3\mid 3^{j-i}$ mean? I thought it meant "such that," but I can't parse it in a way that makes sense with a 3 before it. – LarsH Nov 21 '12 at 16:24
• @LarsH, $a\mid b$ means $b$ is divisible by $a$. Nothing wrong in clarifying doubt. – lab bhattacharjee Nov 21 '12 at 16:26
• Thanks, that clears it up. I also had to look up the notation $(2,3) = 1$. Apparently that's short for $\gcd(2,3) = 1$. – LarsH Nov 21 '12 at 16:50

Laws of GCD:

• $$\gcd(x,y) = \gcd(x,x-y)$$
• for $a$ coprime to $y$: $$\gcd(x,y) = \gcd(x,ay)$$

We can derive general formula using the laws of GCD:

$$\gcd(3^a,2^b) = 1$$

$$\gcd(3^a,2^b-3^a) = 1$$

$$\gcd(3^a,2^{b+c}-2^c 3^a) = 1$$

$$\gcd(3^{a+d},2^{b+c}-2^c 3^a) = 1$$

now put $a+d = s$, $b+c = n$, $a = j-i$, $c = m$ to get the special result.

• thanks! Your answer is very clear! – tlh1987 Nov 21 '12 at 12:44
The text passage contains a mathematical question about the greatest common divisor (GCD) of two expressions and provides a detailed answer explaining why the GCD is 1.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Why the following G.C.D is $1$

Why is the following G.C.D equal to $1$: $$\gcd(3^s, 2^n-3^{(j-i)}2^m),\quad s> j >i \geq 0,$$ and all variables are natural numbers.

• If $i$ ande $j$ are switched it looks right, and easy to show. – coffeemath Nov 21 '12 at 12:19
• @tlh1987 Because $i>j \implies i-j>0 \implies j-i<0 \implies 3^{j-i} \not\in \mathbb{N}$ – badp Nov 21 '12 at 12:22
• @joriki, sorry, there something wrong with the material I have, can I change the question? I am first asked here? – tlh1987 Nov 21 '12 at 12:29
• @badp sorry, I swithed the i and j. – tlh1987 Nov 21 '12 at 12:30
• @tlh1987: When you edit the question in a way that makes an existing answer or comment appear wrong, please mark the edit as such. Thanks. – joriki Nov 21 '12 at 12:55

## 2 Answers

The only prime factor of $3^s$ is 3 as $s\ge 1$

But $2^n-3^{(j-i)}2^m\equiv2^n\pmod 3$ as $3\mid 3^{j-i}$ as $j>i$

So, $2^n-3^{(j-i)}2^m\equiv2^n\equiv(-1)^n\not\equiv 0 \pmod 3$

So, $3^s,2^n-3^{(j-i)}2^m$ can not have any common prime factor, hence $(3^s,2^n-3^{(j-i)}2^m)=1$

• sorry, I switched the i and j, is this still the answer? – tlh1987 Nov 21 '12 at 12:31
• @tlh1987, please find the edited answer. – lab bhattacharjee Nov 21 '12 at 12:33
• Excuse my ignorance... what does the bar in $j>i,3\mid 3^{j-i}$ mean? I thought it meant "such that," but I can't parse it in a way that makes sense with a 3 before it. – LarsH Nov 21 '12 at 16:24
• @LarsH, $a\mid b$ means $b$ is divisible by $a$. Nothing wrong in clarifying doubt. – lab bhattacharjee Nov 21 '12 at 16:26
• Thanks, that clears it up. I also had to look up the notation $(2,3) = 1$. Apparently that's short for $\gcd(2,3) = 1$. – LarsH Nov 21 '12 at 16:50

Laws of GCD:

• $$\gcd(x,y) = \gcd(x,x-y)$$
• for $a$ coprime to $y$: $$\gcd(x,y) = \gcd(x,ay)$$

We can derive general formula using the laws of GCD:

$$\gcd(3^a,2^b) = 1$$

$$\gcd(3^a,2^b-3^a) = 1$$

$$\gcd(3^a,2^{b+c}-2^c 3^a) = 1$$

$$\gcd(3^{a+d},2^{b+c}-2^c 3^a) = 1$$

now put $a+d = s$, $b+c = n$, $a = j-i$, $c = m$ to get the special result.

• thanks! Your answer is very clear! – tlh1987 Nov 21 '12 at 12:44
user
The text passage contains a mathematical question about the greatest common divisor (GCD) of two expressions and provides a detailed answer explaining why the GCD is 1. 

TYPE: SINGLE_QUESTION_WITH_ANSWER