# Find largest square number - without using calculator

• Mar 31st 2011, 11:30 PM
mathguy80
Find largest square number - without using calculator
Hey All,

This problem must be done without the use of a calculator. I am able to guesstimate my way to a close enough answer but need some help to improve this.

If n is the largest square number such that $\displaystyle s \leq n$, find s when

(i) $\displaystyle n = 6.4 \times 10^{3}$

(ii) $\displaystyle n = 6.4 \times 10^{6}$

I got (i) $\displaystyle s = 6400$ easily since its a perfect square,

For (ii) I wrote the number as $\displaystyle 64 \times 10^{4} \times 10$

Then nearest perfect square close to 10 is 9, so $\displaystyle s \approx 8^{2} \times 100^{2} \times 3^{2}$

From there I figured that the number is between $\displaystyle 2400^{2}$ and further got to $\displaystyle s = 2500^{2} = 6250000$

The problem is that the required answer is $\displaystyle 6395841$ and the weight-age is only 1 mark. Using the calculator suggests this is the exactly correct answer.

But since this is a non-calculator use problem, leads me to believe I am missing an obvious/simpler/faster way of getting there. What am I missing?

Thanks again for all your help.
• Apr 1st 2011, 02:31 AM
earboth
Quote:

Originally Posted by mathguy80
Hey All,

This problem must be done without the use of a calculator. I am able to guesstimate my way to a close enough answer but need some help to improve this.

If n <-- typo? Shouldn't that be s? is the largest square number such that $\displaystyle s \leq n$, find s when

(i) $\displaystyle n = 6.4 \times 10^{3}$

(ii) $\displaystyle n = 6.4 \times 10^{6}$

I got (i) $\displaystyle s = 6400$ easily since its a perfect square,

For (ii) I wrote the number as $\displaystyle 64 \times 10^{4} \times 10$

Then nearest perfect square close to 10 is 9, so $\displaystyle s \approx 8^{2} \times 100^{2} \times 3^{2}$

From there I figured that the number is between $\displaystyle 2400^{2}$ and further got to $\displaystyle s = 2500^{2} = 6250000$

The problem is that the required answer is $\displaystyle 6395841$ and the weight-age is only 1 mark. Using the calculator suggests this is the exactly correct answer.

But since this is a non-calculator use problem, leads me to believe I am missing an obvious/simpler/faster way of getting there. What am I missing?

Thanks again for all your help.

I'm only guessing:

$\displaystyle 6.4 \cdot 10^6 = 640 \cdot 100^2$

So $\displaystyle 25 < a < 26$ (see attachment)

Use linear interpolation. According to my sketch $\displaystyle a \approx 25 + \frac{15}{51}$

Since you multiply the apprximate value of a by 100 to get s you only have to calculate the first two digits of $\displaystyle \frac{15}{51} = 0.29$

Thus $\displaystyle a = 25.29~\implies~s=2529$
• Apr 1st 2011, 07:12 AM
mathguy80
Awesome @earboth! A much clearer and very elegant solution. Learning something new everyday here. Thanks again!