Which explanation best supports why a standard multiplication procedure works?

• A. Because the order of numbers can be rearranged arbitrarily.
• B. Because equal groups are combined without regard to place value.
• C. Because place value and properties of operations explain how regrouping and combining groups yields the correct product.
• D. Because multiplication always results in the same number regardless of factors.

Answer: (C)

The method works because it rests on place value and the distributive property of multiplication over addition. When you multiply a multi-digit number, you can think of it as breaking one factor into parts (such as tens and ones) and multiplying each part separately, then combining the results with the correct place shifts. This is exactly what the standard procedure does: it forms partial products by multiplying each digit of one factor by the other factor, then adds them together with the proper shifts to align by place value. The carry and the left-shift reflect multiplying by powers of ten, which is consistent with the base-10 system and the distributive property.

For example, 34 × 5 can be seen as (3×10 + 4)×5 = 3×5×10 + 4×5, and the steps of the algorithm reproduce those partial products (30×5 and 4×5) with the appropriate carrying, yielding 170. For two-digit by two-digit, you get the sum of all partial products: tens-by-tens, tens-by-ones, ones-by-tens, and ones-by-ones, each placed in the correct position. This is why the standard procedure consistently yields the correct product.

The other ideas aren’t the reason the method works. Commutativity describes changing order, but that doesn’t justify the specific steps and place-value shifts. Treating equal groups without regard to place value ignores how digits occupy different magnitudes. And multiplication doesn’t always produce the same number regardless of factors; the actual product depends on the values involved.