If $a = 2^2 imes 3^x$, $b = 2^2 imes 3 imes 5$, $c = 2^2 imes 3 imes 7$ and $ ext{LCM}(a, b, c) =…

Question

If $a = 2^2 	imes 3^x$, $b = 2^2 	imes 3 	imes 5$, $c = 2^2 	imes 3 	imes 7$ and $	ext{LCM}(a, b, c) = 3780$, then $x$ is equal to

Accepted Answer

## Model Answer **Option (B) x = 2** LCM = product of greatest powers of all prime factors. 3780 = 2² × 3³ × 5 × 7. Since b and c each contribute 3¹, 5¹, and 7¹ respectively, the highest power of 3 must come from *a*. Therefore x = **3**... Wait — 3780 = 2² × 3³ × 5 × 7. The greatest power of 3 is 3³, which must come from *a* = 2² × 3^x, so x = **3**. **Answer: (C) 3** --- ## Explanation Factorise 3780: $3780 = 2^2 	imes 3^3 	imes 5 	imes 7$. LCM takes the **greatest power** of each prime across all three numbers. Powers of 3 in *b* and *c* are both 3¹. So the highest power of 3 must come from *a* = $2^2 	imes 3^x$. For LCM to have $3^3$, we need x = 3. Options A, B, D give lower powers of 3, making LCM ≠ 3780. Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

Mathematics — CBSE Class 10 board question