Chapter 13 Solutions — Algebra Play

Let the bottom row of a 4-row pyramid be a, b, c, d (from left to right).

Row 3 (second from bottom): a+b, b+c, c+d

Row 2: (a+b)+(b+c) = a+2b+c, (b+c)+(c+d) = b+2c+d

Row 1 (top): (a+2b+c) + (b+2c+d) = a + 3b + 3c + d

The coefficients 1, 3, 3, 1 match the third row of Pascal's triangle (since the top of an n-row pyramid uses C(n-1,0), C(n-1,1), ..., C(n-1,n-1) as coefficients of the bottom entries).

From the algebraic argument in the chapter: given three digits p < q < r, the largest product is always qp × r — the largest digit is the multiplier, and the remaining two digits form the multiplicand in decreasing order.

Digits: 1 < 3 < 7, so p=1, q=3, r=7.

Largest product = 31 × 7.

Verify by listing all options:

13 × 7 = 91

31 × 7 = 217

17 × 3 = 51

71 × 3 = 213

13 × 7 = 91

37 × 1 = 37

73 × 1 = 73

Maximum is 31 × 7 = 217.

In the trick: start with a 2-digit number ab (digits a and b, b > a). Reverse to get ba. Find the difference ba − ab, then divide by 9.

Using algebra (as shown in the chapter):

ba − ab = (10b + a) − (10a + b) = 9b − 9a = 9(b − a)

Dividing by 9 gives the quotient = b − a.

Relationship: the quotient equals the difference of the two digits of the original number (larger digit minus smaller digit).

Example: start with 47 (a=4, b=7). Difference = 74 − 47 = 27. Quotient = 27 ÷ 9 = 3 = 7 − 4. ✓