Try to prove it!

I saw some formulas regarding HCF and LCM and I am unable to prove them still.

Try if you can, if yes then please write it in the comments

\(LCM(a,b,c)=\frac{a×b×c×HCF(a,b,c)}{HCF(a,b)×HCF(b,c)×HCF(c,a)}\)

HCF(a,b,c)=a×b×c×LCM(a,b,c)LCM(a,b)×LCM(b,c)×LCM(c,a)HCF(a,b,c)=\frac{a×b×c×LCM(a,b,c)}{LCM(a,b)×LCM(b,c)×LCM(c,a)}

LCMoffractions=LCMofnumeratorsHCFofdenominatorsLCM of fractions=\frac{LCM of numerators}{HCF of denominators}

HCFoffractions=HCFofnumeratorsLCMofdenominatorsHCF of fractions=\frac{HCF of numerators}{LCM of denominators}

Try to prove them

#NumberTheory

Note by Sahar Bano
1 year, 1 month ago

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1 vote

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Comments

I can prove the first two formulas. Consider the primes that divide aa, bb , and cc. If we show that the powers of each prime on both sides of the equation are the same, then both sides of the equation are equal. Take some prime pp. Say that pk1ap^{k_{1}}|a, pk2bp^{k_{2}}|b, and pk3cp^{k_{3}}|c, with k1k2k3k_{1}\leq k_{2}\leq k_{3}. Then, the exponent of pp on the left hand side is k3k_{3}. The exponent on the right hand side is k1+k2+k3+k1k1k2k1=k3k_{1}+k_{2}+k_{3}+k_{1}-k_{1}-k_{2}-k_{1}=k_{3}, so both sides are equal. The proof of the second formula is similar.

Advait Nene - 1 year, 1 month ago

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Thanks for the proof

Sahar Bano - 1 year, 1 month ago
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