Question

Find the largest three consecutive natural numbers, such that the sum of one-third of first, one-fourth of second and one-fifth of the third is atmost $25$.

Answer 1

Step 1: Let the required natural numbers be $\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}$ and $\mathbf{(}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{)}$

The required inequality will be, $\frac{1}{3}x+\frac{1}{4}(x+1)+\frac{1}{5}(x+2)\le 25$

On solving the inequality, we get

$$\begin{gathered} \Rightarrow \frac{1}{3}x+\frac{1}{4}x+\frac{1}{5}x\le 25-\frac{1}{4}-\frac{2}{5} \\ \Rightarrow \frac{20x+15x+12x}{60}\le \frac{500-5-8}{20} \\ \Rightarrow \frac{47}{60}x\le \frac{487}{20} \\ \Rightarrow x\le \frac{487}{20}\times \frac{60}{47} \\ \Rightarrow x\le 31 \end{gathered}$$

Thus, the largest value of $x$ is $31$.

Step 2: Evaluate the other two integers

$$\begin{gathered} \Rightarrow x=31 \\ \Rightarrow x+1=32 \\ \Rightarrow x+2=33 \end{gathered}$$

Hence, the largest three consecutive integers are $\mathbf{31}\mathbf{,}\mathbf{}\mathbf{32}$ and $\mathbf{33}$.