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  9. Q32

TNPSC Group-I 2025 · Question 32 of 66

A number x is divisible by 13. When this number is divided by 16, 24, 32 it leaves a remainder 6 in each case. The least value of x is ________.

Aptitude - Number System

  1. A390Correct
  2. B391
  3. C389
  4. D392

Answer: A. 390

Explanation

x = 96k+6 must be divisible by 13; k=4 gives x = 390, the least such value.

Source: TNPSC Group-I Prelims 2025 (held 15 June 2025) - official question paper via tnpscinfo.com (Impacteers TNPSC Coaching compilation)

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