Compute a division with integer and fractional part

In some way you need to estimate $2\sqrt7$. First observe that $2\sqrt7=\sqrt{4\cdot 7}=\sqrt{28}$, thus $5=\sqrt{25}<\sqrt{28}<\sqrt{36}=6$, so

$$4=\left\lfloor\frac{7+5}3\right\rfloor\le\left\lfloor\frac{7+2\sqrt7}3\right\rfloor\le\left\lfloor\frac{7+6}3\right\rfloor=4$$


It is immediate that $2<\sqrt7<3$, which yields

$$\frac{11}3<\frac{7+2\sqrt7}3<\frac{13}3$$ and the answer is one of $3$ or $4$.

So let us evaluate

$$\frac{7+2\sqrt7}3-4=\frac{2\sqrt7-5}3.$$

It is clear that the numerator is positive, by

$$2\sqrt7>5\iff 28>25.$$

Hence, $4$.

You can also start with a tighter bracketing,

$$2.5=\frac52<\sqrt7<3,$$ justified by $$\frac{25}4<7.$$

Then

$$\frac{7+2\cdot\dfrac52}3=4<\frac{7+2\sqrt7}3<\frac{7+2\cdot3}3<5.$$

Tags:

Integers

Fractional Part

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