Practice Westpac maths competition help please!

[ecyrb]

Hey guys.
I have the Westpac Maths Competition tomorrow for year 9 and I have some practice questions from last year that i need help with. If anyone could help me answer these questions I'd be really grateful!

What is the last digit of 6 x 8^2009?

If 3^x = 9^30  then x = ?

Which of the following cannot be the last digit of the sum of the squares of seven consecutive numbers?
a) 3    b) 5    c) 6    d) 7   e) 8

What is the smallest positive integer which, when divided by each of 2, 3, 4, 5, 6 and 7, will give in each case a remainder that is one less than the divisor?

There's probably more

[brightsky]

1. This can be done simply by observing the patterns. Consider the last digits of:













...

Multiplying each of these by six, the pattern doesn't change, so we have we have the pattern: 8, 4, 2, 6, 8, 4, 2, 6, ...

Since

Hence the last digit is 8.

2. , so .

3. Can't think of anything right now, but perhaps method by elimination is suffice.

4.

[ecyrb]

Wow! Thanks for your help.
For the first question I don't really understand how we got the final answer. I understand that the last digits would have the pattern 8,4,2,6 etc. But I don't really understand how we get 4 in the end from the 2009?

And for the 4th question if another question similar to this were to be on this years paper, could i use the same equation for a different set of numbers?
Thanks again!

[/0]

1
when you multiply a number by another, the last digit of the number is unaffected by any of its other digits. Start with 8 and keep multiplying the last digit by 8 and see if you can see a pattern.

2
hint: .

3
Consider the squares from 0 to 9:
0,1,4,9,16,25,36,49,64,81
The last digits are: 0,1,4,9,6,5,6,9,4,1
Now if you start on one of these "last digits" and add the consecutive "last digits" (looping around at the end), you should get 9 sums, and one of the last digits of this sum will not appear as a mult-choice.
(there's might be a faster way to do this)

4.

Let , then will satisfy the property.
But it's not the smallest 'n'
Can you see why?
Consider , why isn't the smallest integer which, when divided by 2 or 4 gives a remainder that is one less than each divisor?

[ecyrb]

:\

Thanks for your response.
I've read the way you solve question 4. over again but i still can't get my head around it haha. I'm pretty slow, is there anyway you can elaborate even more for an idiot like me?

[/0]

2*4 - 1 = 7 works, but 3 is the smallest. This is because 2 and 4 share common factors. What you need to find is the least common multiple of 2,3,4,5,6 and 7.

[ecyrb]

Oh!!!
Ok cool thanks I understand it now
><" took me a while lol.