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Madison
Mathematics
12 June, 15:36
Both parts A and b have three digits multipliers. Why are there three partial products in part a and only two partial products in part b
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Salvatore May
12 June, 16:58
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There are three partial products in part a because both values, 273 and 346, have all numerical values, that aren't 0, so they can be multiplied and a value will be added.
There are only 2 partial products in part b because in 306, the tens place is worth 0 and anything times 0 = 0. There will be nothing to add for that value.
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