System of Equations with GCD

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System of Equations with GCD

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#1 h Mar 10, 2025, 5:07 PM

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Determine the sum of all possible values of $abc$ where $a,b,c$ are positive integers satisfying the equations $\begin{cases} a= \gcd (b,c) + 3 \\ b= \gcd (c,a) + 3 \\ c= \gcd (a,b) + 3 \end{cases}$

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soryn

#2 h Mar 11, 2025, 5:54 AM • 1 Y

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Since gcd of any two possible integers is at least 1,then a>=b>=c>=4(wlog ,since a,b,c>=4). Then,we have: a=gcd(b,c)+3<=c+3. 1) If a=c=>a=b=c=>a=gcd(b,c)+3=a+3, false. 2)If a=c+1 =>b=gcd(a,c)+3=4=> c=4, a=5, contradiction,since a=5, but gcd(b,c)+3=7. 3)If a=c+2, and a is odd=> b=gcd(a,c)+3=4, și c=4, and a=c+2=6, contradiction( a îs odd); if a is even, then b=gcd(a,c)+3=5, and a,c are both even=>c=4,a=6, contradiction,since a=6, but gcd(b,c)+3=4. 4) If a=c+3=> b=4 or b=6. If b=4=>c=4, a=7, and (7,4,4) is a valid solution; if b=6, then gcd(b,c)+3=a>=b=6, obtain gcd(b,c)>=3, and this leads to c=6, a=9 => the valid solution is (9,6,6). . Finally,the answer is 7•4•4+9•6•6=436.

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#3 h Mar 11, 2025, 8:57 AM

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soryn wrote:

Since gcd of any two possible integers is at least 1,then a>=b>=c>=4(wlog ,since a,b,c>=4). Then,we have: a=gcd(b,c)+3<=c+3. 1) If a=c=>a=b=c=>a=gcd(b,c)+3=a+3, false. 2)If a=c+1 =>b=gcd(a,c)+3=4=> c=4, a=5, contradiction,since a=5, but gcd(b,c)+3=7. 3)If a=c+2, and a is odd=> b=gcd(a,c)+3=4, și c=4, and a=c+2=6, contradiction( a îs odd); if a is even, then b=gcd(a,c)+3=5, and a,c are both even=>c=4,a=6, contradiction,since a=6, but gcd(b,c)+3=4. 4) If a=c+3=> b=4 or b=6. If b=4=>c=4, a=7, and (7,4,4) is a valid solution; if b=6, then gcd(b,c)+3=a>=b=6, obtain gcd(b,c)>=3, and this leads to c=6, a=9 => the valid solution is (9,6,6). . Finally,the answer is 7•4•4+9•6•6=436.

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This post has been edited 1 time. Last edited by MrHeccMcHecc, Mar 11, 2025, 8:58 AM

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