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The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 4 in each case, but when divided by 8 leaves no remainder, is
  • a)
    32
  • b)
    64
  • c)
    96
  • d)
    128
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The smallest number, which when divided by 5, 10, 12 and 15, leaves re...
Let the number is N.
N = LCM (5, 10, 12, 15) × x + 4
⇒ N = 60x + 4
⇒ N = 56x + 4x + 4 = (8 × 7x) + 4x + 4
⇒ 4x + 4 is divisible by 8.
⇒ 4 × 1 + 4 = 8 is divisible by 8
⇒ x = 1
∴ N = 60 × 1 + 4 = 64
Alternate method:
Using Hit and trial, 64 when divided by 5, 10, 12 and 15 leaves remainder 4 but when divided by 8 leaves no remainder.
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The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 4 in each case, but when divided by 8 leaves no remainder, isa)32b)64c)96d)128Correct answer is option 'B'. Can you explain this answer?
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