Number system

HCF × LCM

HCF × LCM = a × b

Only for two numbers

Sum of first n natural numbers

n(n + 1)/2

Sum of squares

n(n + 1)(2n + 1)/6

Sum of cubes

[n(n + 1)/2]²

Number of factors

N = pᵃqᵇrᶜ → (a+1)(b+1)(c+1)

Divisibility by 11

(odd-place sum) − (even-place sum) = 0 or 11k

Unit digit cycle

Powers repeat every 4: 2→2,4,8,6 3→3,9,7,1

Sum of first n odd numbers

n²

Sum of first n even numbers

n(n + 1)

Trailing zeros in n!

⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …