Mathematics — Standard (041) — CBSE Class 10 board question

Prime Factorisation:

$26 = 2 \times 13$

$65 = 5 \times 13$

$117 = 3 \times 3 \times 13 = 3^2 \times 13$

HCF = Product of the smallest powers of common prime factors

Common prime factor = 13

$\therefore \text{HCF}(26, 65, 117) = 13$

LCM = Product of the greatest powers of all prime factors

$\therefore \text{LCM}(26, 65, 117) = 2^1 \times 3^2 \times 5^1 \times 13^1 = 2 \times 9 \times 5 \times 13 = \mathbf{1170}$

Source: Chapter 1, Section 1.2 (The Fundamental Theorem of Arithmetic)

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• HCF: Take only the prime factors common to all three numbers, each raised to the lowest power present. Here only 13 is common to all three → HCF = 13.
• LCM: Take every prime factor that appears in any of the numbers, each raised to its highest power → LCM = 2 × 3² × 5 × 13 = 1170.
• Show the factorisation step clearly — examiners award marks for the working, not just the final answer.
• Note: For three numbers, HCF × LCM ≠ product of the three numbers, so don't try to verify using that shortcut.