If a and b are positive integers with no common factor , show that

[a/b] + [2a/b] + [3a/b] + ... + [(b-1)a/b] = (a-1)(b-1)/2,

where [.] denotes the greatest integer function.

Question

If a and b are positive integers with no common factor , show
that
[a/b] + [2a/b] + [3a/b] + + [(b-1)a/b] = (a-1)(b-1)/2,
where [ ] denotes the greatest integer function

Ajanta Trivedi · Accepted Answer

let $S = [\frac{a}{b}] + [\frac{2 a}{b}] + [\frac{3 a}{b}] + .......... [\frac{(b - 1) a}{b}]$ ...........(1)

writing the terms in reverse order:

$S = [\frac{(b - 1) a}{b}] + [\frac{(b - 2) a}{b}] + ........... [\frac{2 a}{b}] + [\frac{a}{b}]$ ..............(2)

adding eq(1) and eq(2):

$2 S = {[\frac{a}{b}] + [\frac{(b - 1) a}{b}]} + {[\frac{2 a}{b}] + [\frac{(b - 2) a}{b}]} + ............. {[\frac{(b - 1) a}{b}] + [\frac{a}{b}]} 2 S = {[\frac{a}{b}] + [a - \frac{a}{b}]} + {[\frac{2 a}{b}] + [a - \frac{2 a}{b}]} + ................ + {[a - \frac{a}{b}] + [\frac{a}{b}]}$ ............(3)

let $\frac{a}{b} = I + f$ where I denotes the integral part and f denotes the fractional part.

therefore

$[\frac{a}{b}] + [a - \frac{a}{b}] = [I + f] + [a - I - f] = I + a - I - 1 = a - 1$

similarly let $[\frac{2 a}{b}] = I_{1} + f_{1}$ where $I_{1} a n d f_{1}$ are the integral and fractional parts respectively.

$[\frac{2 a}{b}] + [a - \frac{2 a}{b}] = [I_{1} + f_{1}] + [a - I_{1} - f_{1}] = I_{1} + a - I_{1} - 1 = a - 1$

similarly all the pairs in eq(3) result a-1.

thus:

$2 S = (a - 1) + (a - 1) + ..................... + (a - 1) {u p t o (b - 1) t e r m s} 2 S = (a - 1) (b - 1) S = \frac{(a - 1) (b - 1)}{2}$

hope this helps you.

cheers!!

If a and b are positive integers with no common factor , show that [a/b] + [2a/b] + [3a/b] + ... + [(b-1)a/b] = (a-1)(b-1)/2, where [.] denotes the greatest integer function.

If a and b are positive integers with no common factor , show that

[a/b] + [2a/b] + [3a/b] + ... + [(b-1)a/b] = (a-1)(b-1)/2,

where [.] denotes the greatest integer function.