Algebra Play · Lesson 6 of 10
The Largest Product
“Optimise digit-placement products by systematic comparison, place-value algebra, and a general proof.”
• List all possible placements of three distinct digits in a two-digit-number times one-digit-number product. • Reduce an exhaustive search by comparing candidates systematically. • Use place value to rewrite a two-digit number algebraically. • Prove the arrangement that gives the largest product when p<q<r. • Distinguish evidence from examples from a general algebraic proof.
The largest-product puzzle asks us to place three digits into a two-digit number multiplied by a one-digit number, using each digit once. Trying a few arrangements may find the answer, but the deeper question is why one arrangement must be best.
Start by Listing Every Possibility
With digits 2, 3, and 5, there are six arrangements: 23×5, 25×3, 32×5, 35×2, 52×3, and 53×2. A systematic list prevents us from accidentally missing a case.
| Multiplier | Two possible multiplicands |
|---|---|
| 2 | 35, 53 |
| 3 | 25, 52 |
| 5 | 23, 32 |
For a fixed multiplier, the larger two-digit multiplicand gives the larger product. Therefore we only need to keep 53×2, 52×3, and 32×5.
Problem
Determine the largest product using each digit once.
- 1.Compare within equal-multiplier pairs and keep 53×2, 52×3, 32×5.
- 2.53×2=106, 52×3=156, and 32×5=160.
- 3.The largest is 32×5=160.
- 4.The largest digit 5 is the multiplier; among the remaining digits, 3 is placed in the tens position.
Use Place Value Instead of Raw Multiplication
A two-digit number with tens digit q and ones digit p is 10q+p. Writing numbers this way lets us compare products symbolically.
The number being multiplied. In 32×5, the two-digit number 32 is the multiplicand and 5 is the multiplier.
Suppose the digits satisfy p<q<r. After comparing pairs with the same multiplier, only three candidates remain: rq×p, rp×q, and qp×r.
Eliminating the First Candidate
Compare rq×p with rp×q. They are (10r+q)p and (10r+p)q. Expanding gives 10rp+pq and 10rq+pq. The second has 10rq instead of 10rp. Since q>p and r is positive, rp×q is larger.
Comparing the Final Two
Now compare rp×q with qp×r. Expanding gives (10r+p)q = 10rq+pq and (10q+p)r = 10qr+pr. The large place-value terms 10rq and 10qr are equal. The decision comes down to pq versus pr. Since r>q and p is non-negative, pr≥pq, and for positive p it is strictly larger.
Problem
For positive distinct digits p<q<r, which arrangement is largest?
- 1.The candidate rq×p is smaller than rp×q.
- 2.Compare rp×q and qp×r.
- 3.Their tens-place contributions are equal: 10rq.
- 4.The remaining terms are pq and pr, and pr>pq because r>q and p>0.
- 5.Therefore qp×r is largest: use r as the multiplier and arrange the other two digits in decreasing order.
Problem
Find the largest product.
- 1.Order the digits: 1<3<7, so p=1, q=3, r=7.
- 2.The rule gives qp×r = 31×7.
- 3.31×7=217.
- 4.A quick comparison with the other likely candidate 71×3=213 confirms the result.
What If One Digit Is Zero?
The textbook-style rule is clearest for positive distinct digits. If p=0, the final comparison pq versus pr gives equality: both are zero. Then rp×q and qp×r can tie. For example, with 0,3,7, 70×3 and 30×7 both equal 210. This shows why stating assumptions matters in a proof.
Do not conclude that the largest-looking two-digit number must be used as the multiplicand. A slightly smaller multiplicand paired with a much larger multiplier can produce the larger product.
Use digits 3,5,9. Predict the largest arrangement from the rule, then calculate only the two strongest candidates to verify it.
Why the General Proof Is More Useful Than a Table
For one set of digits we could simply multiply all six products. The algebraic proof is more valuable because it tells us what to do before any multiplication: sort the positive digits p<q<r, put r as the multiplier, and form qp from the other two. The proof also tells us exactly which assumptions make that rule strict.
This is an optimisation argument. We are not finding an unknown from an equation; we are comparing several permitted choices and proving that one choice cannot be beaten. Algebra supports both solving and comparing.
Quiz
How many arrangements are possible with three distinct digits in a two-digit number times a one-digit number?
For digits 2,3,5, what is the largest product?
If p<q<r, which arrangement is largest for positive digits?
Why are 10rq and 10qr equal?
Why should zero be treated carefully in the general proof?
Practice Problems
- Use digits 3,5,9 to find the largest possible product.
- Use digits 1,4,8 to find the largest possible product and explain the arrangement.
- List all six arrangements for digits 2,6,7, then reduce them to three candidates by equal-multiplier comparison.
- Show algebraically why rp×q is larger than rq×p when p<q<r and all are positive.
- Explain why qp×r beats rp×q by comparing only the non-common terms after expansion.
- Investigate digits 0,4,9 and identify whether the largest product is unique.
Order 3<5<9. Use 53×9=477.
Key Takeaways
• A systematic list guarantees that no digit arrangement is missed. • Comparing cases with the same multiplier quickly removes weaker options. • Place value rewrites a two-digit number such as qp as 10q+p. • For positive distinct digits p<q<r, the largest product is qp×r. • A general proof must state its assumptions; including zero can change a strict maximum into a tie.