Question

HCF of 1248, 1352, 2626 and their steps by using prime factorisation method​

Answer 1

$\underline{ \pink{ Finding \: HCF \: using \: prime \: factorisation : }}$

$Given \: numbers \: 1248, 1352 \: and \: 2626$

1248 = 2×2×2×2×2×3×13 =2⁵ ×3¹×13¹

1352 = 2×2×2×13×13 = 2³ × 13²

2626 = 2¹ × 13¹ × 101¹

$\red{ HCF \: of \: 1248,1352 \:and \: 2626 }\\ = 2 \times 13 \\\green { = 26}$

$\blue{ (Product\: of\: the \: smallest \:power }\\\blue{ of \:each \: common \:prime \:factors \:of }\\\blue{ the \: numbers )}$

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Answer 2

Factor of 1248, 1352 and.2626

• 1248 = 2×2×2×2×2×3×13 = 2⁵ × 3¹ × 13¹
• 1352 = 2×2×2×13×13 = 2³ × 13²
• 2626 = 2 × 13 × 101

HCF of 1248, 1352 and.2626

= 2×13

= 26