HCF x LCM identity
For two numbers, HCF x LCM = their product. Given any three of the four values, find the fourth.
12 and 18: HCF 6, LCM 36; 6 x 36 = 216 = 12 x 18.
Trap: The identity holds for exactly TWO numbers, not three or more.
Rapid Revision · Quantitative Aptitude
Number System
HCF, LCM, remainders and digits: the oldest chapter in the book and still the first section of most tests. A handful of identities does all the work.
The 3-minute recap
If you read nothing else tonight, read these 6 lines.
- HCF x LCM = product of the two numbers.
- Divisibility: 3 and 9 use digit sums; 4 uses last two digits; 8 uses last three; 11 uses alternating sum.
- Unit digits of powers cycle with period at most 4.
- Trailing zeros of n! = floor(n/5) + floor(n/25) + floor(n/125) + ...
- Factor count: write n = p^a x q^b, factors = (a+1)(b+1).
- Sum of 1..n = n(n+1)/2; sum of squares = n(n+1)(2n+1)/6.
Formula sheet
Every formula for number system in one place, each labelled so you know exactly when to reach for it. Screenshot it the night before.
Divisibility quick tests
3 -> digit sum div by 3; 9 -> digit sum div by 9; 11 -> (odd-place sum - even-place sum) div by 11; 4 -> last 2 digits; 8 -> last 3 digits
HCF x LCM
HCF(a,b) x LCM(a,b) = a x b
Only holds for exactly two numbers.
Sum of naturals
1 + 2 + ... + n = n(n+1)/2
Sum of squares / cubes
squares = n(n+1)(2n+1)/6 cubes = [n(n+1)/2]^2
Count of factors
if N = p^a x q^b x r^c then factors = (a+1)(b+1)(c+1)
Units digit cyclicity
powers of a digit repeat every 4 (2,3,7,8) or 1-2 (0,1,5,6,4,9)
Take the exponent mod 4 to find the last digit fast.
Work through the cards
9 cards, each one idea: what it is, a worked example, and the trap to dodge.
Divisibility rules
3 or 9: digit sum divisible by 3 or 9. 4: last two digits. 8: last three digits. 11: (sum of odd-position digits) - (sum of even-position digits) divisible by 11.
There is no remainder in 918/9 because 9+1+8 = 18 is a multiple of 9.
Unit digit of big powers
Unit digits repeat in cycles of at most 4. Find the exponent mod 4 (using 4 when the remainder is 0), then read the cycle.
7^43: cycle 7, 9, 3, 1; 43 mod 4 = 3, so unit digit is 3.
Trap: When exponent mod 4 = 0, use the 4th member of the cycle, not the 0th.
Trailing zeros in factorials
Every zero needs a 5 (2s are plentiful). Zeros in n! = floor(n/5) + floor(n/25) + floor(n/125) + ...
100!: 20 + 4 = 24 trailing zeros.
Counting factors
Prime factorize n = p^a x q^b x r^c; the number of factors is (a+1)(b+1)(c+1).
72 = 2^3 x 3^2: (3+1)(2+1) = 12 factors.
Trap: Do not forget 1 and n itself are counted among the factors.
Remainder arithmetic
Remainders respect multiplication and addition: replace each number by its remainder first, then combine, then reduce again.
(17 x 23) mod 5 = (2 x 3) mod 5 = 1.
HCF and LCM of fractions
HCF of fractions = HCF of numerators / LCM of denominators. LCM = LCM of numerators / HCF of denominators.
HCF of 2/3 and 4/9 = HCF(2,4)/LCM(3,9) = 2/9.
Trap: The two formulas are mirror images; writing LCM/LCM is the standard slip.
Standard sums
1+2+...+n = n(n+1)/2. Squares: n(n+1)(2n+1)/6. Cubes: (n(n+1)/2)^2. Odd numbers 1+3+...+(2n-1) = n^2.
1 to 50: 50 x 51 / 2 = 1275.
Successive division
Numbers leaving remainder r when divided by several divisors: answer = LCM(divisors) x k + r.
Smallest number leaving remainder 3 when divided by 4, 5, 6: LCM 60 + 3 = 63.
Go deeper
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