Find three consecutive largest positive integers such that the sum of one-third of first, one-forth of second and one-fifth of third is at most $20$ .

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Solution

Let the numbers be x-1, x, x+1
$\frac{1}{3} (x - 1)^{\times 20} + \frac{1}{4} (x)^{\times 15} + \frac{1}{5} (x + 1)^{\times 12} \leq 20$
$\frac{20 x - 20 + 15 x + 12 x + 2}{60} \leq 20$
$\Rightarrow 47 x - 18 \leq 1200$
$\Leftarrow 47 x \leq 1218$
$\Rightarrow x \leq 25.9$
So x = 25
x-1 = 24, x+1 = 26
So numbers are 24, 25 ,26

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Find three consecutive largest positive integers such that the sum of one-third of first, one-forth of second and one-fifth of third is at most 20.

Let the numbers be x-1, x, x+113(x−1)×20+14(x)×15+15(x+1)×12≤2020x−20+15x+12x+260≤20⇒47x−18≤1200⇐47x≤1218⇒x≤25.9So x = 25x-1 = 24, x+1 = 26 So numbers are 24, 25 ,26

Find three consecutive largest positive integers such that the sum of one-third of first, one-forth of second and one-fifth of third is at most $20$ .