Number System & Simplification

1BODMAS, and the two pairs that tie

Brackets, Orders, Division, Multiplication, Addition, Subtraction. The acronym is read as six ranks, but it is really four: division and multiplication share a rank, and addition and subtraction share another. Within a shared rank you work strictly left to right, and that single sentence is where most simplification marks are lost — 36 ÷ 6 × 3 is 18, not 2.

The bar drawn over an expression, the vinculum, outranks even brackets: it is the innermost grouping there is. Clear it first, then work outwards through the brackets.

Figure. BODMAS is really an instruction to build this tree and then evaluate it from the leaves back to the root. The two leaves at the far right are one rank, so they are settled before the + they feed; the bracket is settled independently; only then does the − at the root fire. Nothing nearer the root can be evaluated until everything beyond it is a number.

How it works

1. Clear the barAnything under a vinculum is evaluated before the bracket it sits inside.
2. Brackets outwardsInnermost first: ( ), then { }, then [ ]. Never open two at once.
3. Ranks, then left to rightOrders, then ÷ and × together, then + and − together — scanning left to right within each pair.

2Divisibility, and how to test a composite

The rules for 2, 4, 5, 8 and 10 read the tail of the number; the rules for 3 and 9 read its digit sum; the rule for 11 reads the alternating digit sum. For 918082 the odd places give 9 + 8 + 8 = 25 and the even places 1 + 0 + 2 = 3, and 25 − 3 = 22 is a multiple of 11 — so the number is, and indeed 918082 = 11 × 83462.

For a composite divisor there is no single rule, and the correct move is to split it into factors that are coprime. 72 = 8 × 9 works because 8 and 9 share no factor. 12 = 2 × 6 does not: 18 passes both those tests and is not a multiple of 12.

No diagram — every test reads digits: the tail for 2, 4, 5, 8 and 10, the digit sum for 3 and 9, the alternating sum for 11, coprime factors for a composite. The rules table beside this concept is the visual.

3Unit digits, and the cycle of four

The unit digit of a power depends on nothing but the unit digit of the base, and it repeats with a period of at most 4. Powers of 7 run 7, 9, 3, 1 and then start again; powers of 2 run 2, 4, 8, 6; powers of 4 and 9 have period 2; and 0, 1, 5 and 6 never change at all.

So reduce the exponent modulo 4 and read the cycle. The one thing to watch is a remainder of 0: an exponent that is an exact multiple of 4 lands on the fourth entry of the cycle, not the first.

Figure. Multiplying by 7 once more walks you one step round this loop, and after four steps you are back where you started. The exponent's remainder on division by 4 names the node — with the fourth node carrying remainder 0, which is the only place a candidate ever comes off this diagram wrong.

How it works

1. Throw the base awayKeep only its unit digit; 17, 27 and 1237 all behave exactly like 7.
2. Reduce the exponent mod 4Divide the exponent by 4 and keep the remainder, treating a remainder of 0 as 4.
3. Read the cycleCount that many places along the base's cycle, starting from the first power.

4Remainders, and the identity behind them

Dividend = Divisor × Quotient + Remainder, with the remainder always at least 0 and always less than the divisor. That identity is the whole subject: any statement about a remainder is a statement that a number has this form, and most remainder questions are solved by writing the number in that form and then re-dividing.

Remainders multiply and add just as the numbers do, provided you reduce again at the end. And a remainder may be taken negatively when that is smaller: 29 leaves 4 on division by 5, but treating it as −1 is quicker and gives the same answer.

Figure. The ticks are the multiples of 5, and 17 sits between two of them. The distance forward from 15 is the ordinary remainder, 2. The distance back from 20 is the negative remainder, −3, and the two always sum to the divisor. Either may be used in a calculation; pick whichever is smaller in size and convert at the end.

How it works

1. Write the number out"Leaves 47 on division by 342" means the number is 342q + 47 for some whole q.
2. Split by the new divisorIf the new divisor divides 342, the whole 342q term vanishes and only 47 survives.
3. Reduce what is left47 is still bigger than 19, so divide once more: 47 = 19 × 2 + 9.

5Prime factorisation, and counting the factors

Write N = pᵃ qᵇ rᶜ. Any factor is built by choosing an exponent for each prime independently — anything from 0 to a for p, and so on — so the number of factors is (a+1)(b+1)(c+1). The same independence gives the sum of all the factors as a product of geometric series, one per prime.

Two consequences are worth carrying. Factors pair off around √N, so the count is odd exactly when N is a perfect square and its square root is left unpaired. And the number of ways of writing N as a product of two factors is therefore half the count, rounded up for a square.

Figure. For 360=2^3\cdot 3^2\cdot 5^1 each prime's exponent is chosen independently: 4 choices, 3 choices and 2 choices, so (3+1)(2+1)(1+1)=24 factors. The bars are the choice counts, not the primes.

How it works

1. Factorise completelyStrip out 2s first, then 3s, then 5s; stop when the quotient is prime.
2. Add one to each exponentThe extra choice is the exponent 0, which means leaving that prime out of the factor.
3. Multiply the choicesEach combination of exponents is one distinct factor, so the counts multiply.

6Trailing zeroes in a factorial

A trailing zero is a factor of 10, and a factor of 10 is a 2 paired with a 5. Inside n! the 2s vastly outnumber the 5s — every second number contributes a 2 but only every fifth contributes a 5 — so the number of zeroes is simply the number of 5s available.

Count them in a ladder: ⌊n/5⌋ picks up one 5 from each multiple of 5, ⌊n/25⌋ picks up the second 5 that every multiple of 25 secretly carries, ⌊n/125⌋ the third, and so on until the term is zero. Forgetting the higher rungs is the only mistake this question ever tests.

Figure. Each floor division peels off another power of 5. Summing \lfloor 250/5\rfloor+\lfloor 250/25\rfloor+\lfloor 250/125\rfloor=62 counts the trailing zeroes in 250! — the 2s never run out first.

7The standard sums

Four results carry almost every series question at this level: 1 + 2 + ⋯ + n = n(n+1)/2, the sum of squares n(n+1)(2n+1)/6, the sum of cubes which is the square of the first result, and the sum of the first n odd numbers, which is exactly n².

None needs memorising blind. The first is the count of cells in a staircase, which is half a rectangle n by n+1; the third says the cubes add up to the square of the triangular number, so 1³ + 2³ + ⋯ + 10³ = 55² = 3025. What these are really for is subtraction: a sum over part of a range is the sum to the top minus the sum below the bottom.

Figure. The staircase holds 1 + 2 + 3 + 4 cells and sits inside a block 4 rows tall and 5 columns wide. The cells it leaves empty form the same staircase upside down, so the staircase is exactly half the block — which is n(n+1)/2, drawn rather than recalled. The +1 in the formula is the one extra column the staircase needs to be mirrored into.

8Rationals, irrationals and recurring decimals

A rational number is one that can be written p/q with q ≠ 0, and its decimal either stops or repeats forever in a block. An irrational cannot, and its decimal never settles into a block — √2, π and e are the standard examples, and 22/7 is not one of them however often it stands in for π.

Which rationals stop? Put the fraction in lowest terms and look at the denominator: if its only prime factors are 2 and 5 the decimal terminates, because you can scale it into a power of ten. 7/40 terminates at 0.175; 7/30 cannot, because of the 3.

Going the other way, a repeating block of k digits becomes that block over k nines, and digits before the block are handled by putting the same count of zeroes after the nines and subtracting off the non-repeating part.

Figure. The split at the root is total: every real number is on one side or the other, and the decimal expansion decides which. Integers are rationals with denominator 1, so they sit on the rational branch and not beside it — a question asking whether an integer is rational is asking whether you have read this tree correctly.