If $a = 2^3 \cdot 3^4 \cdot 5^2$ and $b = 2^4 \cdot 3^2 \cdot 7$, then $\dfrac{\text{LCM}(a,b)}{\text{HCF}(a,b)}$ equals
A$2^7 \cdot 3^6 \cdot 5^2 \cdot 7$
B$2 \cdot 3^2 \cdot 5 \cdot 7$
C$2 \cdot 3^2 \cdot 5^2 \cdot 7$
D$2^4 \cdot 3^4 \cdot 5^2 \cdot 7$
Answer & Solution
Correct answer: C. $2 \cdot 3^2 \cdot 5^2 \cdot 7$
**HCF takes min powers:** $2^3 3^2$.
**LCM takes max powers:** $2^4 3^4 5^2 7$.
Ratio $= 2^{4-3} 3^{4-2} 5^2 7 = 2\cdot3^2\cdot5^2\cdot7$.
Keep practising Unit 1B2: Algebraic Limits and Continuity
Browse all Unit 1B2: Algebraic Limits and Continuity questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
The decimal expansion of $\dfrac{257}{2^3 \cdot 5^4}$ terminates after how many decimal plThe product of the HCF and LCM of 6 and 20 is:Which of the following is an irrational number?The remainder when $2^{100}$ is divided by 7 isHow many positive integers less than 1000 are divisible by neither 2 nor 5?The HCF of two consecutive positive integers isThe smallest positive integer that is divisible by every integer from 1 to 10 isThe largest number that divides 245 and 1029, leaving remainders 5 and 5 respectively, is