Numerical example for $gcd(a,b)=prod p_i^min(a_i,b_i)$

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I'm actually having trouble understanding the above corollary. Can anyone please provide a numerical example of that corollary?
Thank You So Very Much in advance.




Corollary If $a=prod p_i^a_i$ and $b=prod p_i^b_i$ where the $p_i$ are distinct primes, then $gcd(a,b)=prod p_i^min(a_i,b_i)$. The LCM is $prod p_i^max(a_i,b_i)$ and the product of the last two expressions is $ab$.











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    I'm actually having trouble understanding the above corollary. Can anyone please provide a numerical example of that corollary?
    Thank You So Very Much in advance.




    Corollary If $a=prod p_i^a_i$ and $b=prod p_i^b_i$ where the $p_i$ are distinct primes, then $gcd(a,b)=prod p_i^min(a_i,b_i)$. The LCM is $prod p_i^max(a_i,b_i)$ and the product of the last two expressions is $ab$.











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      $begingroup$


      I'm actually having trouble understanding the above corollary. Can anyone please provide a numerical example of that corollary?
      Thank You So Very Much in advance.




      Corollary If $a=prod p_i^a_i$ and $b=prod p_i^b_i$ where the $p_i$ are distinct primes, then $gcd(a,b)=prod p_i^min(a_i,b_i)$. The LCM is $prod p_i^max(a_i,b_i)$ and the product of the last two expressions is $ab$.











      share|cite|improve this question











      $endgroup$




      I'm actually having trouble understanding the above corollary. Can anyone please provide a numerical example of that corollary?
      Thank You So Very Much in advance.




      Corollary If $a=prod p_i^a_i$ and $b=prod p_i^b_i$ where the $p_i$ are distinct primes, then $gcd(a,b)=prod p_i^min(a_i,b_i)$. The LCM is $prod p_i^max(a_i,b_i)$ and the product of the last two expressions is $ab$.








      elementary-number-theory greatest-common-divisor prime-factorization least-common-multiple






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      edited Feb 17 at 16:58









      Martin Sleziak

      44.8k10119273




      44.8k10119273










      asked Feb 17 at 14:20









      Jessi JhaJessi Jha

      233




      233




















          3 Answers
          3






          active

          oldest

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          5












          $begingroup$

          Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
          beginalign*
          gcd(48,40)
          &=gcd(a,b) \
          &= prod p_i^min(a_i,b_i) \
          &= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
          &= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
          &= 2^3cdot 3^0 cdot 5^0 \
          &= 8,
          endalign*



          Using a similar method we determine that



          beginalign*
          lcm(48,40)
          &=lcm(a,b) \
          &= prod p_i^max(a_i,b_i) \
          &= 2^4cdot 3^1 cdot 5^1 \
          &= 240.
          endalign*



          We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$



          Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Formula for LCM should use Max and not minumum. The calculations are right sl
            $endgroup$
            – Edcookie274
            Feb 17 at 15:23











          • $begingroup$
            Thanks for pointing that out. Fixed.
            $endgroup$
            – Ben
            Feb 17 at 15:32






          • 1




            $begingroup$
            Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:18


















          5












          $begingroup$

          $$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
          b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
          textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
          textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
          textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
          endalign$$






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank You so much!!!! All these examples really helped me!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:19


















          3












          $begingroup$

          For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank You so much!!! I got it! !
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:21










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          3 Answers
          3






          active

          oldest

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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
          beginalign*
          gcd(48,40)
          &=gcd(a,b) \
          &= prod p_i^min(a_i,b_i) \
          &= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
          &= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
          &= 2^3cdot 3^0 cdot 5^0 \
          &= 8,
          endalign*



          Using a similar method we determine that



          beginalign*
          lcm(48,40)
          &=lcm(a,b) \
          &= prod p_i^max(a_i,b_i) \
          &= 2^4cdot 3^1 cdot 5^1 \
          &= 240.
          endalign*



          We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$



          Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Formula for LCM should use Max and not minumum. The calculations are right sl
            $endgroup$
            – Edcookie274
            Feb 17 at 15:23











          • $begingroup$
            Thanks for pointing that out. Fixed.
            $endgroup$
            – Ben
            Feb 17 at 15:32






          • 1




            $begingroup$
            Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:18















          5












          $begingroup$

          Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
          beginalign*
          gcd(48,40)
          &=gcd(a,b) \
          &= prod p_i^min(a_i,b_i) \
          &= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
          &= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
          &= 2^3cdot 3^0 cdot 5^0 \
          &= 8,
          endalign*



          Using a similar method we determine that



          beginalign*
          lcm(48,40)
          &=lcm(a,b) \
          &= prod p_i^max(a_i,b_i) \
          &= 2^4cdot 3^1 cdot 5^1 \
          &= 240.
          endalign*



          We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$



          Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Formula for LCM should use Max and not minumum. The calculations are right sl
            $endgroup$
            – Edcookie274
            Feb 17 at 15:23











          • $begingroup$
            Thanks for pointing that out. Fixed.
            $endgroup$
            – Ben
            Feb 17 at 15:32






          • 1




            $begingroup$
            Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:18













          5












          5








          5





          $begingroup$

          Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
          beginalign*
          gcd(48,40)
          &=gcd(a,b) \
          &= prod p_i^min(a_i,b_i) \
          &= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
          &= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
          &= 2^3cdot 3^0 cdot 5^0 \
          &= 8,
          endalign*



          Using a similar method we determine that



          beginalign*
          lcm(48,40)
          &=lcm(a,b) \
          &= prod p_i^max(a_i,b_i) \
          &= 2^4cdot 3^1 cdot 5^1 \
          &= 240.
          endalign*



          We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$



          Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).






          share|cite|improve this answer











          $endgroup$



          Certainly: consider $a=48=2^4cdot 3$ and $b=40=2^3cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that
          beginalign*
          gcd(48,40)
          &=gcd(a,b) \
          &= prod p_i^min(a_i,b_i) \
          &= p_1^min(a_1,b_1)cdot p_2^min(a_2,b_2)cdot p_3^min(a_3,b_3) \
          &= 2^min(4,3)cdot 3^min(1,0)cdot 5^min(0,1) \
          &= 2^3cdot 3^0 cdot 5^0 \
          &= 8,
          endalign*



          Using a similar method we determine that



          beginalign*
          lcm(48,40)
          &=lcm(a,b) \
          &= prod p_i^max(a_i,b_i) \
          &= 2^4cdot 3^1 cdot 5^1 \
          &= 240.
          endalign*



          We can then observe that $$ab=48cdot 40=1920=8cdot 140=gcd(48,40)cdot lcm(48,40)=gcd(a,b)cdot lcm(a,b).$$



          Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 17 at 15:31

























          answered Feb 17 at 14:34









          BenBen

          329110




          329110











          • $begingroup$
            Formula for LCM should use Max and not minumum. The calculations are right sl
            $endgroup$
            – Edcookie274
            Feb 17 at 15:23











          • $begingroup$
            Thanks for pointing that out. Fixed.
            $endgroup$
            – Ben
            Feb 17 at 15:32






          • 1




            $begingroup$
            Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:18
















          • $begingroup$
            Formula for LCM should use Max and not minumum. The calculations are right sl
            $endgroup$
            – Edcookie274
            Feb 17 at 15:23











          • $begingroup$
            Thanks for pointing that out. Fixed.
            $endgroup$
            – Ben
            Feb 17 at 15:32






          • 1




            $begingroup$
            Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:18















          $begingroup$
          Formula for LCM should use Max and not minumum. The calculations are right sl
          $endgroup$
          – Edcookie274
          Feb 17 at 15:23





          $begingroup$
          Formula for LCM should use Max and not minumum. The calculations are right sl
          $endgroup$
          – Edcookie274
          Feb 17 at 15:23













          $begingroup$
          Thanks for pointing that out. Fixed.
          $endgroup$
          – Ben
          Feb 17 at 15:32




          $begingroup$
          Thanks for pointing that out. Fixed.
          $endgroup$
          – Ben
          Feb 17 at 15:32




          1




          1




          $begingroup$
          Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
          $endgroup$
          – Jessi Jha
          Feb 20 at 3:18




          $begingroup$
          Thank You so much!!!! Your answer gave me complete clarity! !!! Thank you once again!!!
          $endgroup$
          – Jessi Jha
          Feb 20 at 3:18











          5












          $begingroup$

          $$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
          b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
          textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
          textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
          textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
          endalign$$






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank You so much!!!! All these examples really helped me!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:19















          5












          $begingroup$

          $$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
          b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
          textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
          textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
          textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
          endalign$$






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank You so much!!!! All these examples really helped me!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:19













          5












          5








          5





          $begingroup$

          $$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
          b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
          textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
          textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
          textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
          endalign$$






          share|cite|improve this answer











          $endgroup$



          $$beginaligna=600&=2^3cdot 3^1 cdot 5^2\
          b=54&=2^1cdot 3^3 cdot 5^0\[2ex]
          textGCD(a,b)&=2^1cdot 3^1cdot 5^0=6\
          textLCM(a,b)&=2^3cdot 3^3cdot 5^2=5400\[2ex]
          textGCD(a,b)cdottextLCM(a,b)&=2^4cdot 3^4cdot 5^2=32400=ab
          endalign$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 17 at 14:43

























          answered Feb 17 at 14:32









          Daniel MathiasDaniel Mathias

          1,36018




          1,36018











          • $begingroup$
            Thank You so much!!!! All these examples really helped me!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:19
















          • $begingroup$
            Thank You so much!!!! All these examples really helped me!!
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:19















          $begingroup$
          Thank You so much!!!! All these examples really helped me!!
          $endgroup$
          – Jessi Jha
          Feb 20 at 3:19




          $begingroup$
          Thank You so much!!!! All these examples really helped me!!
          $endgroup$
          – Jessi Jha
          Feb 20 at 3:19











          3












          $begingroup$

          For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank You so much!!! I got it! !
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:21















          3












          $begingroup$

          For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank You so much!!! I got it! !
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:21













          3












          3








          3





          $begingroup$

          For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.






          share|cite|improve this answer









          $endgroup$



          For example, $540 = 2^2cdot 3^3cdot 5$ and $72 = 2^3cdot 3^2$, so their GCD is $2^2cdot 3^2 = 36$ and their LCM is $2^3cdot 3^3cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 17 at 14:32









          rogerlrogerl

          18k22748




          18k22748











          • $begingroup$
            Thank You so much!!! I got it! !
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:21
















          • $begingroup$
            Thank You so much!!! I got it! !
            $endgroup$
            – Jessi Jha
            Feb 20 at 3:21















          $begingroup$
          Thank You so much!!! I got it! !
          $endgroup$
          – Jessi Jha
          Feb 20 at 3:21




          $begingroup$
          Thank You so much!!! I got it! !
          $endgroup$
          – Jessi Jha
          Feb 20 at 3:21

















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