    Question

# If a=pl.qm.rn and b=px.qy.rz where p,q,r are primes and p<q<r and l,m,n,x,y,z are all natural numbers. Given HCF = p2.q3.r and LCM = p3.q4.r3, which one of the following is NOT true?

A

l + x = 5

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B

m + y = 7

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C

n + z = 4

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D

l + m + n = x + y + z

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Solution

## The correct option is A l + m + n = x + y + z HCF = The least powers of the common factors Here we have HCF = p2.q3.r p2=P(min(l,x)), so either l = 2 or x = 2 q3=q(min(m,y)), so either m = 3 or y = 3 r=r(min(n,z)), so either n = 1 or z = 1 LCM = The highest power of all the prime factors Similarly, we have LCM = p3 q4 r3 p3=p(max(l,x)), either l = 3 or x = 3 q4=q(max(m,n)), either m = 4 or y = 4 r3=r(max(n,z)), either n = 3 or z = 3 So from the above l = 2 or 3 and x = 2 or 3 so l + x = 2 + 3 = 5 Similarly m= 3 or m = 4 and y = 3 or y = 4 So m + y = 3 + 4 = 7 Similarly either n = 3 or z = 3 or n = 1 or z = 1 So n + z = 3 + 1 = 4 So from the options : 1) x+l=2+3=5 Option 1 is true 2) m+y=3+4=7 Option 2 is true 3) n+z=3+1=4 Option 3 is true 4) l+m+n≠x+y+z ( If we take any values for l ,x etc ) Option 4 is the answer  Suggest Corrections  0      Similar questions  Related Videos   The Fundamental Theorem of Arithmetic
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