What is the smallest number which when divided by 64 and 80 leaves a remainder of 9 in each case?
Given:
We need to find the smallest number that, when divided by 64 and 80, leaves a remainder of 9 in each case.
Formula Used:
Required number = LCM of divisors + Remainder
Solution:
Divisors are 64 and 80.
Prime factorization:
64 = 26, 80 = 24 × 5
LCM = 26 × 5 = 320
Required number = LCM + Remainder
=> Required number = 320 + 9
=> Required number = 329
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