Question
Download Solution PDFWhat is the smallest number which when divided by 64 and 80 leaves remainder 9 in each case?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
We need to find the smallest number which, when divided by 64 and 80, leaves a remainder of 9 in each case.
Formula used:
Required number = LCM of divisors + Remainder
Calculation:
Divisors are 64 and 80. Find their LCM:
Prime factorization:
64 = 26, 80 = 24 × 5
LCM = 26 × 5 = 320
Required number = LCM + Remainder
⇒ Required number = 320 + 9
⇒ Required number = 329
∴ The correct answer is option (3).
Last updated on Jul 8, 2025
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