find the smallest number which when divided by 15,20,48 will in each case leave 9 as the reminder
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Answer:
Factors of
15 = 3x5
20 = 2x2x5
48 = 2x2x2x2x3
LCM = 2x2x2x2x3x5 = 240
240+9 = 249
The answer is 249.
249 divided by 15 or 20 or 48 leaves a remainder of 9 in each case.
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