Question

If $a=p^l.q^m.r^n$ and $b=p^x.q^y.r^z$ where p,q,r are primes and p<q<r and l,m,n,x,y,z are all natural numbers. Given HCF = $p^2$.$q^3.r$ and LCM = $p^3$.$q^4$.$r^3$, which one of the following is NOT true?

• A. l + x = 5
• B. m + y = 7
• C. n + z = 4
• D. l + m + n = x + y + z

Answer 1

The correct option is D

l + m + n = x + y + z

HCF = The least powers of the common factors

Here we have HCF = $p^2.q^3.r$

$p^2=P^{\left(min\right(l,x\left)\right)}$, so either l = 2 or x = 2

$q^3=q^{\left(min\right(m,y\left)\right)}$, so either m = 3 or y = 3

$r=r^{\left(min\right(n,z\left)\right)}$, so either n = 1 or z = 1

LCM = The highest power of all the prime factors

Similarly, we have LCM = p3 q4 r3

$p^3=p^{\left(max\right(l,x\left)\right)}$, either l = 3 or x = 3

$q^4=q^{\left(max\right(m,n\left)\right)}$, either m = 4 or y = 4

$r^3=r^{\left(max\right(n,z\left)\right)}$, 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) $l+m+n\ne x+y+z$ ( If we take any values for l ,x etc )

Option 4 is the answer