# Year 8 maths. quick

• Jun 27th 2008, 04:54 AM
claudeluvstwilight
Year 8 maths. quick
This is again urgent maths hw. please help me. thx

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?
• Jun 27th 2008, 05:18 AM
topsquark
Quote:

Originally Posted by claudeluvstwilight
This is again urgent maths hw. please help me. thx

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

$693 = 3^2 \cdot 7 \cdot 11$

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 $4 = 2^2$.

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
• Jun 27th 2008, 05:18 AM
Moo
Hello (Wink)

Quote:

Originally Posted by claudeluvstwilight
This is again urgent maths hw. please help me. thx

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

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

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.

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

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

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

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

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

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

Is it clear enough ? (Worried)
• Jun 27th 2008, 05:19 AM
topsquark
Quote:

Originally Posted by Moo
Hello (Wink)

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

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.

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

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

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

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

But 9 is 3 squared. :)

-Dan
• Jun 27th 2008, 05:30 AM
Moo
Quote:

Originally Posted by topsquark
But 9 is 3 squared. :)

-Dan

I edited some things in my rush (Tongueout)
• Jul 25th 2008, 07:53 PM
mathswonderer
Intereted in this one, does that mean tht 308 isthe only possible answer? How do we know for sure(Thinking) 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