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$(m,n)=(4,-20)$ works too.

jasperE3 wrote: Find all positive integer solutions of the equation $10(m+n)=mn$.

Math4Life2020 wrote: Use SFFT: $(m-5)(n-5)=25$. Solutions: m=6, n=30 m=10, n=10 m=30, n=6 Actually, using SFFT, you get (m-10)(n-10) = 100.

Math4Life2020 wrote: Use SFFT: $(m-5)(n-5)=25$. Solutions: m=6, n=30 m=10, n=10 m=30, n=6 Plugging this in doesn't work... I have: $(m,n)=(11,110),(12,60),(14,35),(15,30),(20,20),(30,15),(35,14),(60,12),(110,11),$ by using @above's SFFT factoring.

How is this 1st grade?

aops-g5-gethsemanea2 wrote: How is this 1st grade? Bro, if 1st graders knew that in croatia, then all of them would be in AIME in 4th

Alex-131 wrote: aops-g5-gethsemanea2 wrote: How is this 1st grade? Bro, if 1st graders knew that in croatia, then all of them would be in AIME in 4th I'm putting my bets on 2nd grade in AIME, and then 4th grade in USAMO.

6th grade MOP, 7th TST and IMO

1st grade refers to 1st grade of high school I believe.

Apple321 wrote: 6th grade MOP, 7th TST and IMO Pre K AIME, Kindergarten USAJMO, MOP, 1st Grade TST and IMO

Iamnobody wrote: Math4Life2020 wrote: Use SFFT: $(m-5)(n-5)=25$. Solutions: m=6, n=30 m=10, n=10 m=30, n=6 Plugging this in doesn't work... I have: $(m,n)=(11,110),(12,60),(14,35),(15,30),(20,20),(30,15),(35,14),(60,12),(110,11),$ by using @above's SFFT factoring. Sorry , I made a mistake.

kante314 wrote: Apple321 wrote: 6th grade MOP, 7th TST and IMO Pre K AIME, Kindergarten USAJMO, MOP, 1st Grade TST and IMO koff Luke koff koff koff phew am I having a fit koff

Use Simon's Favorite Factoring Trick. $$mn-10m-10n=0 \Longrightarrow m(n-10)-10n+100=100 \Longrightarrow (m-10)(n-10)=100$$. Thus finding all the factors of $100$ which are $1, 2, 4, 5, 10, 20, 25, 50, 100$, we can find the answer.