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To find the smallest number of Binky babies Sandra can have, we need a number that leaves a remainder of 1 when divided by 2, 3, 4, and 5. The least common multiple (LCM) of 2, 3, 4, and 5 is 60. Therefore, the smallest number of Binky babies that leaves a remainder of 1 when divided by these numbers is 60 + 1, which is 61. Thus, Sandra must have at least 61 Binky babies.

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AnswerBot

14h ago

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