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Question Number 87586 by manr last updated on 05/Apr/20

l.c.m of two numbers is p^2 q^4 r^4  p q r are  primes.find the possible no. of pairs

$${l}.{c}.{m}\:{of}\:{two}\:{numbers}\:{is}\:{p}^{\mathrm{2}} {q}^{\mathrm{4}} {r}^{\mathrm{4}} \:{p}\:{q}\:{r}\:{are} \\ $$$${primes}.{find}\:{the}\:{possible}\:{no}.\:{of}\:{pairs} \\ $$

Answered by mr W last updated on 05/Apr/20

lcm(x,y)=p^2 q^4 r^4   say x=p^i q^j r^k  with i,j,k≥0  say y=p^a q^b r^c  with a,b,c≥0  max(i,a)=2 ⇒2×3=6 possiblities  max(j,b)=4 ⇒2×5=10 possiblities  max(k,c)=4 ⇒2×5=10 possiblities  ⇒6×10×10=600 possible pairs for x, y.

$${lcm}\left({x},{y}\right)={p}^{\mathrm{2}} {q}^{\mathrm{4}} {r}^{\mathrm{4}} \\ $$$${say}\:{x}={p}^{{i}} {q}^{{j}} {r}^{{k}} \:{with}\:{i},{j},{k}\geqslant\mathrm{0} \\ $$$${say}\:{y}={p}^{{a}} {q}^{{b}} {r}^{{c}} \:{with}\:{a},{b},{c}\geqslant\mathrm{0} \\ $$$${max}\left({i},{a}\right)=\mathrm{2}\:\Rightarrow\mathrm{2}×\mathrm{3}=\mathrm{6}\:{possiblities} \\ $$$${max}\left({j},{b}\right)=\mathrm{4}\:\Rightarrow\mathrm{2}×\mathrm{5}=\mathrm{10}\:{possiblities} \\ $$$${max}\left({k},{c}\right)=\mathrm{4}\:\Rightarrow\mathrm{2}×\mathrm{5}=\mathrm{10}\:{possiblities} \\ $$$$\Rightarrow\mathrm{6}×\mathrm{10}×\mathrm{10}=\mathrm{600}\:{possible}\:{pairs}\:{for}\:{x},\:{y}. \\ $$

Commented by mr W last updated on 05/Apr/20

MJS sir: is this correct?

$${MJS}\:{sir}:\:{is}\:{this}\:{correct}? \\ $$

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