# Thread: Year 8 maths. quick

1. ## Year 8 maths. quick

Find all three-digit even numbers N such that 693 x N is a perfect square, that is, 693 x N = k to the power of 2 where k is an integer

i am in yr 8. please explain in easy words...:P but answer it completly (i know im mean but wat u have dun will work) thx!

i get it and how it works but how did u get 693 = the factors 7,3,11. how did u get those 3 numbers?

2. Originally Posted by claudeluvstwilight

Find all three-digit even numbers N such that 693 x N is a perfect square, that is, 693 x N = k to the power of 2 where k is an integer
$\displaystyle 693 = 3^2 \cdot 7 \cdot 11$

So any number we multiply this by to make a perfect square has to contain factors of 7 and 11. Since this number has to be even as well, it needs to contain a $\displaystyle 4 = 2^2$.

The smallest N that qualifies is 308. I leave it to you to find the rest (if any).

Hint: What is the smallest number you can multiply this by and still leave the product a perfect square?

-Dan

3. Hello

Originally Posted by claudeluvstwilight

Find all three-digit even numbers N such that 693 x N is a perfect square, that is, 693 x N = k to the power of 2 where k is an integer
$\displaystyle 693=63*11=7*9*11=7^1*3^2*11^1.$

A perfect square is formed by the product of numbers to an even power.

So here, you have to multiply 693 by a number that will yield even powers of 7, 3, 11 in the product.

$\displaystyle 7^1 \rightarrow \text{multiply by {\color{red}7}} \rightarrow 7^2$

$\displaystyle 3^2 \rightarrow \text{OK.}$

$\displaystyle 11^1 \rightarrow \text{multiply by {\color{red}11}} \rightarrow 11^2$

--> $\displaystyle N=7*11=77$

Multiply N by all perfect squares, until you don't have 3-digit numbers anymore.

$\displaystyle 4*77=\boxed{308} \rightarrow \text{OK.}$
$\displaystyle 9*77=\boxed{693} \rightarrow \text{OK.}$
$\displaystyle 16*77=1232 \rightarrow \text{not OK.}$

Is it clear enough ?

4. Originally Posted by Moo
Hello

$\displaystyle 693=63*11=7*9*11=7^1*3^2*11^1.$

A perfect square is formed by the product of numbers to an even power.

So here, you have to multiply 693 by a number that will yield even powers of 7, 9, 11 in the product.

$\displaystyle 7^1 \rightarrow \text{multiply by {\color{red}7}} \rightarrow 7^2$

$\displaystyle 3^2 \rightarrow \text{OK.}$

$\displaystyle 11^1 \rightarrow \text{multiply by {\color{red}11}} \rightarrow 11^2$

--> $\displaystyle N=7*11=77$
But 9 is 3 squared.

-Dan

5. Originally Posted by topsquark
But 9 is 3 squared.

-Dan
I edited some things in my rush

6. Intereted in this one, does that mean tht 308 isthe only possible answer? How do we know for sure because it has to be even.

wasn't it stated that it must be divsible by 3 and to get perfect square you must times all the number by the same thing. E.g. 11 sqauredx 7 squared x 3 squared?

im confused