Methods and worked examples: Algebra Play

2.1 Pyramids with unknown boxes

Sometimes the bottom row is unknown and the top is given. Take the chapter's pyramid with top 60, second row 32 and 28, bottom row 12, c, 8. The rules give three equations: 12 + c = 32 style relations, c + 8 = the other middle number, and the two middle numbers add to 60.

Using letters: 12 + c = a and c + 8 = b, with a + b = 60. Adding the first two equations gives (12 + c) + (c + 8) = a + b = 60, so 20 + 2c = 60, 2c = 40 and c = 20. Then 12 + 20 = 32 and 20 + 8 = 28. The unknown box is found by adding the two equations in which it appears.

For a four-row pyramid with bottom row a, b, c, d, the top is a + 3b + 3c + d. The coefficients 1, 3, 3, 1 come from adding upward twice: the middle numbers are used more often than the end numbers. A five-row pyramid would give 1, 4, 6, 4, 1.

2.2 Calendar magic

A 2 × 2 block from a calendar page has four numbers. If the top-left number is a, the other three become a plus 1, a plus 7 and a plus 8, because the next column adds 1 and the next row adds 7. The sum is 4a + 16.

If a friend reports the sum as 36, solve 4a + 16 = 36: subtract 16 to get 4a = 20, then divide by 4 to get a = 5. The block is 5, 6, 12, 13. The same method works for a block of a different shape once you write each cell as a letter plus offsets; the sum gives an equation for the top-left cell.

2.3 Algebra grids

In an algebra grid, shapes stand for numbers, and the last entry of each row is the sum of the earlier entries. Replace each shape with a letter, write one equation per row, and solve the equations from the simplest row upward. The chapter's grid with row sums 27, 19 and 19 is solved this way: the simplest row fixes one shape, the next row gives a second shape, and the first row then gives the third.

The method is the same as the pyramid method: turn each visual rule into an equation, then substitute values already found. Draw the grid and label every shape before writing anything.

The same place-value idea handles three-digit numbers: 100a + 10b + c, so any pattern in the digits becomes an expression in a, b and c. When two such expressions are equal for every allowed digit, their coefficients must match, which is often faster than testing numerical examples.

2.4 The largest product

Fill three given digits into a two-digit number times a one-digit number, using each digit once. With digits 2, 3, 5 there are six arrangements. List them first as 23 × 5, then 25 × 3, then 32 × 5, then 35 × 2, then 52 × 3, and finally 53 × 2. Group the arrangements by the multiplier: with 2 we get 35 × 2 and 53 × 2; with 3 we get 25 × 3 and 52 × 3; and inside the last group we see 23 × 5 beside 32 × 5. Within a group, the larger multiplicand always gives the larger product, so only three candidates remain: 53 × 2, 52 × 3 and 32 × 5.

Now 52 × 3 is clearly bigger than 53 × 2, leaving 52 × 3 against 32 × 5. Expand both: 32 × 5 = 30 × 5 + 2 × 5 and 52 × 3 = 50 × 3 + 2 × 3. The first parts are equal (both 150), and the second part is larger for 32 × 5 (10 against 6). So 32 × 5 = 160 is the largest.

The general proof uses digits p < q < r. In each multiplier pair the larger multiplicand wins, leaving rq × p, rp × q and qp × r. Since q > p, rp × q beats rq × p, and expanding the last two shows that qp × r has the larger second term. The largest digit becomes the multiplier and the other two digits are written in decreasing order.

2.5 Divisibility tricks

A two-digit number ab means 10a + b. Reversing it gives 10b + a. Their difference is 9(b − a), which is always a multiple of 9, so dividing by 9 leaves no remainder and the quotient is b − a, the difference between the digits.

Their sum is 11(a + b), always a multiple of 11. A three-digit number abc can be cycled into bca and cab; the sum is 111(a + b + c) = 3 × 37 × (a + b + c), which is divisible by 37 and by 3. Finally, abcabc = 1001 × abc, and 7 × 11 × 13 = 1001, so dividing by 7, then 11, then 13 returns the original three-digit number. Each trick is an identity in disguise. A quick substitution into the finished formula catches sign errors before they spread to the next step.