An algebra problem by Vatsalya Tandon

Algebra Level 4

⌊ x 10 ⌋ + ⌊ x 20 ⌋ + ⌊ x 30 ⌋ = 11 x 60 \large\left \lfloor \dfrac x{10} \right \rfloor + \left \lfloor \dfrac x{20} \right \rfloor + \left \lfloor \dfrac x{30} \right \rfloor = \dfrac{11x}{60}

Find the number of solutions of x x in the interval 0 < x < 1000 0<x<1000 satisfying the equation above.

Notation : ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 16.

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Vatsalya Tandon by Vatsalya Tandon , 21, India

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1 solution

Vatsalya Tandon
Jul 12, 2016

11 x 60 = x 10 + x 20 + x 30 \large \frac{11x}{60} = \frac{x}{10} + \frac{x}{20} + \frac{x}{30}

∴ \therefore { x 10 \large \frac{x}{10} } + { x 20 \large \frac{x}{20} } + { x 30 \large \frac{x}{30} } = 0

Where {.} denotes fractional part, and since it can never be negative-

{ x 10 \large \frac{x}{10} } = { x 20 \large \frac{x}{20} } = { x 30 \large \frac{x}{30} } = 0

x x is a multiple of 60

Number of multiples of 60 60 from 0 0 to 1000 1000 = 16 16

Ans - 16 \large \boxed{16}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Didnt understand

Mr Yovan - 4 years, 11 months ago

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