Lessons / Quantitative aptitude
Quantitative aptitude · Day 1

Number System

Most number questions never need the full number. Divisibility wants the last few digits, unit digit wants the last digit, remainders want small leftovers. Throw away everything the question does not ask about.

Practise (19 questions) Present as slidesCheat sheet: Summations and seriesCheat sheet: Squares, cubes and powersCheat sheet: Unit digits of powers (cyclicity)

Why this topic matters

Most number questions never need the whole number. 'Is it divisible by 8?' needs the last 3 digits. 'What is the unit digit?' needs the last digit. Learn which part to look at.

Number system is the most common quant topic in placement papers: divisibility, unit digit, remainders, factors, HCF and LCM. Each one has a rule that avoids big calculations, and those rules are also used inside every other chapter.

Kinds of numbers

Natural numbers: 1, 2, 3 ... Whole numbers: 0, 1, 2, 3 ... Integers add the negatives: ... -2, -1, 0, 1, 2 ...

A prime has exactly two factors, 1 and itself: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ... 1 is NOT prime, and 2 is the only even prime. A composite number has more than two factors (4, 6, 8, 9 ...).

To test if a number up to 400 is prime, divide only by primes up to its square root. For 221: √(221) is about 14.9, so try 2, 3, 5, 7, 11, 13. 13 × 17 = 221, so it is not prime.

Co-primes are two numbers whose only common factor is 1, like 8 and 15. They do not have to be prime themselves.

Primes below 100 (25 of them)
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Example 1Is 391 a prime number?
  1. √(391) is just under 20, so test primes up to 19.If 391 had a factor bigger than 20, it would also have one smaller than 20.
  2. 2, 3, 5, 7, 11, 13: none divides. 17 × 23 = 391.Check each prime in turn.
  3. Not prime: 391 = 17 × 23.

Divisibility rules

You can tell whether a number divides another without dividing. Each rule looks at only part of the number.

2, 5, 10: look at the last digit. 4: last two digits. 8: last three digits. 3 and 9: add all the digits. 11: alternate plus and minus across the digits; the result must be 0 or a multiple of 11.

For a composite divisor, split it into co-prime parts: divisible by 12 means divisible by 3 AND 4; by 72 means by 8 AND 9.

DivisorTestExample
2last digit even3,758
3digit sum divisible by 34,212 (sum 9)
4last two digits divisible by 47,316 (16)
5last digit 0 or 58,275
6divisible by 2 and 35,334
8last three digits divisible by 891,128 (128)
9digit sum divisible by 96,705 (sum 18)
11alternating sum 0 or multiple of 118,261 (8-2+6-1 = 11)
12divisible by 3 and 44,332
7double the last digit, subtract from the rest; repeat343 → 34 - 6 = 28
Example 1Find the smallest digit * so that 5*2 is divisible by 6.
  1. Divisible by 6 means divisible by 2 and by 3.6 = 2 × 3, and 2 and 3 are co-prime.
  2. It ends in 2, so it is already even.
  3. Digit sum 5 + * + 2 = 7 + * must be divisible by 3, so * = 2, 5 or 8.
  4. Smallest is 2.
Example 2Is 7,29,432 divisible by 72?
  1. 72 = 8 × 9, co-prime parts.Check both rules; both must pass.
  2. Last three digits 432 / 8 = 54, yes.Rule for 8.
  3. Digit sum 7+2+9+4+3+2 = 27, divisible by 9, yes.Rule for 9.
  4. Yes.

Unit digit of big powers

The last digit of a product depends only on the last digits of the numbers. And the last digits of powers repeat in a cycle.

2, 3, 7, 8 repeat every 4 powers. 4 and 9 repeat every 2. 0, 1, 5, 6 never change.

Method: divide the power by 4 and look at the remainder r. The unit digit is the same as for the power r (if r = 0, use power 4).

7^n7n=19n=23n=31n=4
Unit digits of 7n repeat every 4: 7, 9, 3, 1
2^n2n=14n=28n=36n=4
Unit digits of 2n repeat every 4: 2, 4, 8, 6
Example 1Unit digit of 795.
  1. 7's cycle: 7, 9, 3, 1 (length 4).71 = 7, 72 = 49, 73 = 343, 74 = 2401, then it repeats.
  2. 95 = 4 × 23 + 3, remainder 3.Full cycles do not change the position.
  3. Third in the cycle: 3.
Example 2Unit digit of 253 × 362.
  1. 53 = 4 × 13 + 1, so 253 ends like 21 = 2.
  2. 62 = 4 × 15 + 2, so 362 ends like 32 = 9.
  3. 2 × 9 = 18, unit digit 8.
Exam tipWrite the four-cycles of 2, 3, 7 and 8 on your rough sheet once. Every unit-digit question is then 10 seconds.

Factors, and how many a number has

Factorise into primes first, using a factor tree. 360 = 23 × 32 × 5.

Number of factors: add 1 to each power and multiply. For 360 that is (3+1)(2+1)(1+1) = 24.

Why: a factor of 360 picks 2 zero, one, two or three times (4 choices), 3 zero, one or two times (3 choices), and 5 zero or one time (2 choices).

360218029024531535
360 = 2 × 2 × 2 × 3 × 3 × 5
Example 1How many factors does 720 have?
  1. 720 = 24 × 32 × 5.Prime factorisation.
  2. (4+1)(2+1)(1+1) = 5 × 3 × 2 = 30.Add one to each power, multiply.

HCF and LCM

HCF (highest common factor) is the biggest number that divides all of them. LCM (lowest common multiple) is the smallest number all of them divide.

From prime factors: HCF takes the LOWEST power of each common prime; LCM takes the HIGHEST power of every prime present.

For two numbers only: HCF × LCM = product of the numbers.

Word clues: 'largest', 'greatest', 'maximum size of tiles' → HCF. 'Smallest', 'least', 'when will they meet again', 'bells ring together' → LCM.

HCF × LCM = a × b
Two numbers only.
Largest number dividing a, b, c leaving the same remainder
HCF of the differences (b - a), (c - b).
Smallest number leaving remainder r when divided by x, y, z
LCM(x, y, z) + r.
HCF of fractions = HCF of numerators / LCM of denominators
LCM of fractions = LCM of numerators / HCF of denominators.
Example 1Three bells ring every 12, 15 and 20 minutes. They ring together at 9:00. When do they next ring together?
  1. They ring together at common multiples of 12, 15 and 20.'Together again' is an LCM clue.
  2. LCM(12, 15, 20) = 60 minutes.12 = 22 × 3, 15 = 3 × 5, 20 = 22 × 5 → 22 × 3 × 5.
  3. At 10:00.
Example 2Find the greatest number that divides 43, 91 and 183 leaving the same remainder.
  1. Differences: 91 - 43 = 48, 183 - 91 = 92.If both leave remainder r, the remainder cancels in the difference.
  2. HCF(48, 92) = 4.
  3. 4.

Remainders

The remainder of a sum or product can be found from the remainders of the parts. 17 × 23 / 12: 17 leaves 5 and 23 leaves 11, 5 × 11 = 55, which leaves 7.

For big powers, find a small power that leaves remainder 1, then use it. 23 = 8 leaves 1 when divided by 7, so 299 = (23)33 leaves 1 too.

Negative remainders are allowed and often faster: 23 leaves -1 when divided by 12.

Example 1Remainder when 2100 is divided by 7.
  1. 23 = 8 = 7 + 1 leaves remainder 1.Look for a power just above a multiple of 7.
  2. 2100 = (23)33 × 2.100 = 3 × 33 + 1.
  3. Remainder = 133 × 2 = 2.Each 23 contributes a remainder of 1.
Example 2Remainder when 2525 is divided by 26.
  1. 25 = 26 - 1 leaves remainder -1.A negative remainder is often easier.
  2. (-1)25 = -1.Odd power keeps the minus sign.
  3. -1 means 26 - 1 = 25.

Trailing zeros of n!

A zero at the end comes from a factor 10 = 2 × 5. In n! there are always more 2s than 5s, so count the 5s.

Zeros in n! = n/5 + n/25 + n/125 + ... (drop the remainders each time).

Example 1How many zeros does 100! end with?
  1. 100/5 = 20.Every multiple of 5 gives one 5.
  2. 100/25 = 4.Multiples of 25 give a second 5.
  3. 20 + 4 = 24.

Formula summary

Divisible by 3 / 9
Digit sum divisible by 3 / 9.
By 4 / 8
Last 2 / last 3 digits divisible by 4 / 8.
By 11
(sum of odd-place digits) - (sum of even-place digits) is 0 or a multiple of 11.
By 7
Double the last digit and subtract it from the rest; repeat. 343 → 34 - 6 = 28, divisible.
Unit digit cycles
2: 2,4,8,6. 3: 3,9,7,1. 7: 7,9,3,1. 8: 8,4,2,6. 4 and 9 cycle in 2. 0,1,5,6 never change. Power mod 4, remainder 0 = last in the cycle.
Remainder theorem
Rem(a × b) = Rem(a) × Rem(b), reduced again. Find a power that leaves remainder 1 and use it to kill big exponents.
HCF × LCM = a × b
Two numbers only.
HCF / LCM of fractions
HCF = HCF(numerators)/LCM(denominators). LCM = LCM(numerators)/HCF(denominators).
Trailing zeros of n!
n/5 + n/25 + n/125 ... (whole parts only).
Same remainder r for all divisors
Number = k × LCM + r.

Shortcuts and tricks

Greatest / smallest fraction without decimals

Use it when: Four fractions in the options.

  1. Cross-multiply two at a time: a/b vs c/d → compare a*d with b*c.
  2. If numerators rise by the same step and denominators by the same step, compare that step fraction to the first: if step fraction is bigger, the list is rising.
ExampleLargest of 3/5, 5/8, 7/11, 9/14
  1. Numerators +2, denominators +3. Step fraction 2/3 = 0.67.
  2. 2/3 is bigger than 3/5, so each next fraction is bigger. Last one wins.

Remainder of a huge power

Use it when: 2100 / 7 style.

  1. Find the smallest power leaving remainder 1 (23 = 8 → remainder 1).
  2. Write the exponent as multiples of that plus leftover: 2100 = (23)33 × 2.
  3. (1)33 × 2 = 2.
ExampleRemainder of 2100 / 7
  1. 23 leaves 1.
  2. 299 leaves 1, times 2.

Common mistakes

Calling 1 a prime.
A prime has exactly two factors. 1 has one.
Using HCF × LCM = product for three numbers.
It works for two numbers only.
Unit digit cycle: using the power itself instead of its remainder by 4.
795 → 95 mod 4 = 3 → third in the cycle.

Before you move on, you should be able to...

  • test a number for 3, 4, 8, 9 and 11 without dividing
  • find the unit digit of a big power from its cycle
  • count the factors of a number from its prime powers
  • choose HCF or LCM from the words in the question
  • find the remainder of a big power

Practice questions (19)

Try each one before opening the solution. Or practise them one by one so your score is saved.

easy Least number that leaves remainder 4 when divided by 6, 9 and 15
  1. LCM(6, 9, 15) = 90.
  2. 90 + 4 = 94.

Answer: 94

easy HCF of two numbers is 16, LCM 480. One number is 96. Find the other.
  1. 16 × 480 / 96 = 80.

Answer: 80

easy What is the unit digit of 843?
  1. 8's cycle: 8, 4, 2, 6.Length 4.
  2. 43 = 4 × 10 + 3.Remainder 3.
  3. Third in the cycle: 2.

Answer: 2

easy How many factors does 180 have?
  1. 180 = 22 × 32 × 5.Prime factors.
  2. (2+1)(2+1)(1+1) = 18.

Answer: 18

easy How many zeros are at the end of 60!?
  1. 60/5 = 12, 60/25 = 2.Count the 5s.
  2. 12 + 2 = 14.

Answer: 14

easy Find the least number which when divided by 12, 16 and 24 leaves remainder 5 each time.
  1. LCM(12, 16, 24) = 48.Divisible by all three.
  2. 48 + 5 = 53.

Answer: 53

medium Unit digit of 795 × 358
  1. 95 mod 4 = 3 → 7's cycle → 3.
  2. 58 mod 4 = 2 → 3's cycle → 9.
  3. 3 × 9 = 27 → 7.

Answer: 7

medium Remainder when 17 × 23 × 29 is divided by 12
  1. 17 → 5, 23 → 11, 29 → 5.
  2. 5 × 11 = 55 → 7. 7 × 5 = 35 → 11.

Answer: 11

medium Largest 4-digit number divisible by 12, 15 and 18
  1. LCM = 180.
  2. 9999 / 180 = 55 rem 99.
  3. 9999 - 99 = 9900.

Answer: 9900

medium Number of trailing zeros in 250!
  1. 250/5 = 50, /25 = 10, /125 = 2.
  2. 50 + 10 + 2 = 62.

Answer: 62

hard Greatest number that divides 1657 and 2037 leaving remainders 6 and 5
  1. Subtract the remainders: 1651 and 2032.
  2. HCF(1651, 2032) = 127.

Answer: 127

hard Find the digit * so that 73*54 is divisible by 11
  1. Alternate sum: 7 - 3 + * - 5 + 4 = 3 + *.
  2. It must be 0 or 11, so * = 8.

Answer: 8

hard Remainder when 3250 is divided by 13
  1. 33 = 27 leaves 1.
  2. 250 = 3 × 83 + 1.
  3. 1 × 31 = 3.

Answer: 3

hard Remainder when 7100 is divided by 13
  1. 13 is prime, so 712 leaves 1 (Fermat).
  2. 100 = 12 × 8 + 4, so 7100 leaves the same as 74 = 2401.
  3. 2401 mod 13 = 9.

Answer: 9

hard How many factors does 1800 have?
  1. 1800 = 23 × 32 × 52.
  2. (3+1)(2+1)(2+1) = 36.

Answer: 36

hard Sum of all factors of 72
  1. 72 = 23 × 32.
  2. (1+2+4+8)(1+3+9) = 15 × 13.

Answer: 195

hard Highest power of 7 that divides 1000!
  1. 1000/7 = 142, /49 = 20, /343 = 2.
  2. 142 + 20 + 2.

Answer: 164

hard Smallest number that leaves remainder 3 when divided by 5, 6, 7, 8 but is divisible by 9
  1. Form 840k + 3.
  2. 840 leaves 3 mod 9, so 3k + 3 must be divisible by 9 → k = 2.
  3. 1683.

Answer: 1683

hard Unit digit of 22026 × 32027 × 72028
  1. 2026 mod 4 = 2 → 4.
  2. 2027 mod 4 = 3 → 7.
  3. 2028 mod 4 = 0 → 1.
  4. 4 × 7 × 1 = 28 → 8.

Answer: 8

Class plan (2 hours, for trainers)

A tested order for teaching this topic in one 2-hour session. Present mode follows the same order.

10 minSpeed maths warm-up.
10 minKinds of numbers; test 221 and 391 for primes together.
20 minDivisibility table; find-the-missing-digit questions in pairs.
15 minUnit digits: the cycles on the board.
15 minFactors and the factor tree; number of factors.
20 minHCF and LCM: word clues, bells, tiles, remainder types.
10 minRemainders of products and powers.
5 minTrailing zeros.
10 minPractice set, timed.
5 minRecap and exit ticket: unit digit of 347 (answer 7).
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