SSC GD Constable 6 Feb 2025 Shift 1 Paper

Section: Elementary Mathematics

Question : 41 of 80

Marks: +1, -0

The greatest number, which divides 406 and 1388 to leave 1 and 3 respectively as remainders, is:

Solution:

Required greatest number = HCF of (406-1) and (1388-3)
Calculation:

406-1=405

1388-3=1385

\;\text " ⇒ Find "\; {\HCF}(405,1385)

Prime factorization:

\;405=3 × 3 × 3 × 3 × 5

\;1385=5 × 277

Common factor

=5

⇒ HCF=5

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