13.5 The Largest Product

Fill the digits 4, 6, 7 into \boxed{\ \ }\,\boxed{\ \ } \times \boxed{\ \ }, a two-digit number times a one-digit number, each digit used once. What is the largest product possible?

There are six arrangements:

46\times 7,\quad 47\times 6,\quad 64\times 7,\quad 67\times 4,\quad 74\times 6,\quad 76\times 4.

Group them by the single multiplier. In each pair, the larger two-digit number wins, so we need only compare three:

76\times 4 = 304, \qquad 74\times 6 = 444, \qquad 64\times 7 = 448.

The biggest is 64\times 7 = \mathbf{448}. Notice what won: the largest digit became the one-digit multiplier, and the remaining two digits formed the two-digit number in decreasing order.

Does the pattern always hold?

Let the three digits be p < q < r. Comparing the strong candidates qp \times r and rp \times q by expanding place value:

qp \times r = (10q + p)\,r = 10qr + pr, \qquad rp \times q = (10r + p)\,q = 10rq + pq.

The first terms 10qr and 10rq are equal. The second terms are pr and pq, and since r > q we have pr > pq. So qp \times r wins. Algebra confirms the rule:

The Rule To maximise (two-digit) \times (one-digit) from three given digits, use the largest digit as the single multiplier, and lay the other two in decreasing order as the two-digit number.

Figure it Out: Largest Product

Practice

  1. Fill the digits 2, 4, 9 into \boxed{\ }\,\boxed{\ } \times \boxed{\ } to make the largest product.
  2. Fill the digits 4, 6, 8 into \boxed{\ }\,\boxed{\ } \times \boxed{\ } to make the largest product.

Use the rule: largest digit as multiplier, other two in decreasing order.

  1. Largest digit 9 as multiplier; other two 4, 2 → 42. Product = 42 \times 9 = \mathbf{378}.
  2. Largest digit 8 as multiplier; other two 6, 4 → 64. Product = 64 \times 8 = \mathbf{512}.