Missing Number

What the exam asks. A few numbers are written as a set, such as (14, 280, 5), as a pair joined by an arrow, 25 → 100, or with a bracket, 3 (13) 4. One small formula joins the numbers inside every complete set. You are asked in one of three ways: find the ?, select the option set that follows the same relation, or select the odd pair out.

The logic behind it. The paper-setter first chooses a formula and then feeds numbers into it. So the result is always built from the inputs, and its size tells you how. A result close to the inputs points to a sum or a difference. A result many times bigger points to a product, a square or a power. Look at the size of the result and at what it divides by before you try any formula.

Method.
1. Divide the biggest number of a set by each of the others. A whole quotient that repeats in every set shows multiplication. Here 280 ÷ 14 = 20 = 5 × 4 and 336 ÷ 7 = 48 = 12 × 4, so the middle number is first × last × 4.

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2. If nothing divides, write the sum, difference and product of the inputs and compare each with the result: equal, double, or plus or minus a small number. For the pairs in the next figure, 5 + 3 and 5 × 3 do not match 34, but 5² + 3² = 34 and 6² + 2² = 40.

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3. Still nothing? Look for squares, cubes and powers.
4. Confirm on a second set. A formula that fits only one set is a coincidence.

Triplet and bracket forms. In (a, b, c) any one number can be the result, so ask which of the three is the biggest. The pattern can also be a chain: (7, 41, 211) and (12, 66, 336) follow "multiply by 5 and add 6" twice. In a bracket a (b) c, the middle usually holds a × c, a + c, a² ± c or c² − a².

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Power forms. If a number looks like a big perfect power, break it up: 4913 = 17³ and 6561 = 9⁴, so (17, 4913, 3) and (9, 6561, 4) both follow (a, aᶜ, c).

Odd pair out. Find the formula that three pairs obey; the fourth breaks it. In 11, 121; 14, 196; 22, 484; 18, 300 the second number is the square of the first in three pairs, but 18² = 324, not 300.

What the exam asks. A series of five or six numbers has one gap. The gap can be at the end (4, 6, 8, 10, 12, ?), but in recent papers it is often in the middle: 7, 11, 19, 31, ?, 67. Three related forms sit in the same family: a set with one wrong term, a series of powers, and a growth chain (each term some percent more than the one before).

The logic behind it. A series is made by one pattern that runs from term to term. The steps between neighbours reveal that pattern, so first look at how fast the numbers grow. Equal steps mean an arithmetic series. Steps that themselves grow by a fixed amount mean a squares-type series. Very fast growth, or steps that look random, hint at powers or multiplication.

Method: the difference ladder. Write the differences of neighbouring terms in a row below, then the differences of those, until a row becomes constant or shows a clear pattern. Extend that row by one step and rebuild the rows above it.

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Gap in the middle. You get free checking. Build the ladder from the side that has more terms, fill the gap, then confirm that the next printed term fits the next step. In the figure, 47 + 20 = 67 closes the loop.

Wrong term in a Fibonacci-like set. Every term from the third should be the sum of the two before it (or three, if the question says so). Test from the left; the first failing term is the suspect. Correct it and check that the next term now agrees. For 4, 5, 9, 14, 24, 37, we get 9 + 14 = 23, not 24, and 14 + 23 = 37 confirms that 24 is wrong.

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Powers. Split every term into base and power and study the bases. In 8, 27, 125, 343 the bases 2, 3, 5, 7 are primes, so the next term is 11³ = 1331. A number can be a power in two ways (64 = 8² = 4³), so decide from the whole series.

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Growth chain. If the ratio of neighbours is equal (not the difference), each term is a fixed percent more than the previous one. 4000 growing 10% each step gives 4400, 4840, 5324. The first value is the base, so the nth term uses n − 1 steps.

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What the exam asks. A 3 × 3 matrix (sometimes 3 × 4 or 4 × 3) is shown with one cell marked ?. In the latest papers it is usually the last cell of the last row, as in 11, 13, 143. Some items put letters instead of numbers, and a new form asks which rows or columns of a big table are related to each other.

The logic behind it. A matrix is a stack of small formulas. Either every row obeys one formula or every column does, never both in the same item. In each line the biggest number is usually the result and the other numbers are the inputs. Notice where the big number sits. If the third number of every row is far bigger than the first two, the formula runs along the rows.

Method.
1. Choose the direction: rows or columns.
2. On the first complete line, divide the result by the inputs, then try sum, product, product ± c and squares.
3. Test the guess on the second line. Two lines usually leave only one pattern.
4. Apply it to the incomplete line. If the ? is an input, write the pattern as an equation and solve it backwards.

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For 4, 5, 21 and 3, 6, 19 the pattern is third = first × second + 1. For 7, ?, 15 we get 7 × ? + 1 = 15, so ? = 2.

Column pattern. If the numbers of a row look unrelated but the columns look tidy, run the same method down the columns. In a 4-row grid the pattern may use three numbers of a column. Digits written side by side (3, 4 giving 34) are the last idea to try.

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Letters in a matrix. Write the alphabet position beside each letter (A = 1 … Z = 26), solve it as a number matrix, and convert the answer back. Opposite letters (A–Z, B–Y …) add to 27.

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[New Pattern] Related rows or columns. A larger table is given and you must pick the lines that are connected. Write each column as a formula in the row number n or in another column, and pick those which are tied together by a simple one-step formula. In the PYQ below, Column II = (Column I)² + 2 and Column IV = Column I × (Column I − 1).

What the exam asks. Three triangles, circles or pyramids are drawn side by side and one number is replaced by ?. Because the item is a picture, it seldom appears as a text question, but it belongs to the same family as the matrix: the figure is the unit. The pattern lives inside one figure, and you read it from the complete figures.

The logic behind it. The setter puts the numbers on fixed places of a figure (top, left, right; sectors; blocks) and joins them by a formula. So place matters: you must fix a reading order and never change it. Complete figure 1 gives the guess, complete figure 2 proves it, figure 3 gets the answer.

Triangle. Fix the order top, left, right. For (3, 2, 4) with centre 10 and (4, 3, 5) with centre 17, the sum fails (9, 12) and the product is too big. Product of two plus the third works: 3 × 2 + 4 = 10 and 4 × 3 + 5 = 17. So centre = top × left + right, and the third triangle (5, 4, 6) gives 26. If the centre is much bigger than the sum of the outer numbers, expect a product or a square.

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Circle with sectors. Test the total first: add the outer numbers and compare with the centre (equal, half, double). Outer 3, 5, 7, 9 gives 24 with centre 12, and 2, 4, 6, 8 gives 20 with centre 10, so centre = half the total. For a centre 14 the outer numbers must add to 28; with 5, 6, 7 known, the missing number is 28 − 18 = 10. If the total fails, compare opposite sectors (equal sums, or one double the other), then try products. In joined circles a shared number counts in both circles.

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Pyramid. Each block is made from the two blocks directly below it, usually as their sum (sometimes the average). Bottom 2, 5, 3, 6 gives 7, 8, 9, then 15, 17, then the top 32. If a lower block is unknown, call it b, write the top as an expression in b and solve. Check the pattern on a printed block before filling the rest.

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Cross, star, Y-shapes and constant-total figures are in the next type.

What the exam asks. Several figures (squares, boxes, crosses, stars, Y-shapes) carry a few numbers each, and one number, or sometimes a letter, is replaced by ?. Like the other figure items, these come as pictures, so there are few text versions. The figure is the unit.

The logic behind it. There are three ways in which a setter can tie the numbers of a figure together, and they differ in how much work they need.
1. Constant total. Every figure has the same sum (or the same product).
2. Letter from the sum. The sum of the numbers in a figure is the alphabet position of the letter printed with it.
3. Centre from the arms. The centre is a function of the arms (cross, star, diamond, Y).

Always test the cheapest idea first: the total of one complete figure.

Constant total. Square 1: 12 + 9 + 11 + 8 = 40; square 2: 15 + 6 + 10 + 9 = 40. So every square totals 40, and the third square, with 7, 16 and 8 known (31), misses 40 − 31 = 9. Confirm on a second figure before you believe it. If the totals differ, test the product: a constant product of 60, 120 or 360 is common, and the missing number is the target ÷ the product of the known numbers.

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Letter from the sum. Box 1: 5 + 6 + 4 = 15 and O is the 15th letter. Box 2: 9 + 8 + 4 = 21 and U is the 21st. In box 3 the letter T gives a target of 20; 7 + 6 = 13 is known, so the missing number is 7. Every box has its own target, set by its own letter. The anchors E = 5, J = 10, O = 15, T = 20, Y = 25 let you count any letter quickly.

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Cross, star and diamond. Name the arms T, B, L, R and pair T with B and L with R. Test the plain sum, T + B − L − R, T × B − L × R and T × B + L × R. Cross 1 (7, 3, 4, 2) gives 13 and cross 2 gives 3 by T × B − L × R, so cross 3 gives 6 × 5 − 4 × 3 = 18.

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Y-shapes have three arms and no pairing. Arms 2, 6, 4 give 24 and 3, 4, 5 give 30: the centre is half the product of the arms, so 4, 5, 6 give 60.

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If ? is an arm, write the pattern with the known numbers and solve it as a small equation.

What the exam asks. There are two arithmetic forms in this chapter, and neither needs a hidden pattern.
1. Equation fit. An equation with letters is printed, for example A × B ÷ C − D + E = 14, and each option gives a value for every letter. You choose the option that makes the equation true.
2. Missing digits. Some digits of an addition, subtraction or multiplication are hidden behind letters, and you find them, then combine them (A + B or A × B).

The logic behind it. In both forms the answer is forced by plain arithmetic, so the work is only to organise it. In equation fit you do not search for a pattern; you test options in the correct order of operations (BODMAS: division and multiplication from left to right, then addition and subtraction). In missing digits every column gives a small equation with one unknown, and you solve the columns from the right.

Equation fit: method.
1. Look at the division first. In A × B ÷ C the product A × B must be exactly divisible by C. When all the other terms are whole numbers and the right side is whole, a fraction can never cancel, so any option where C does not divide A × B is out at once.
2. Evaluate each surviving option in BODMAS order.
3. Stop at the option that gives the right side exactly.

For the equation above: 13 × 3 = 39 and 4 × 13 = 52 are not divisible by 4 and 3, so options (a) and (c) go. Then 4 × 3 ÷ 2 − 5 + 13 = 6 − 5 + 13 = 14 fits.

Missing digits: method. Start at the units column and move left. Write the carry above the next column so that you neither forget it nor use it twice. For 3 A 8 + 2 7 B = 6 0 5: 8 + B ends in 5, so B = 7 with carry 1; A + 7 + 1 ends in 0, so A = 2. For subtraction, turn it into an addition (823 − 454 = 369 means 369 + 454 = 823). For multiplication, the units digit of the product depends only on the units digits of the two numbers.

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Finally put the digits back and check the whole sum.