Started 5th two weeks ago, already being caught off by Honours Maths

TheOwl12

My teacher hasn't explained much to us, but is already giving out detentions to people who have nothing done. I just need help with this homework and tomorrow I'm going to ask him more about the questions as he has just been going off of his own slides and not the book

heres a pic of the q's h.ttp://i.imgur.com/oWp5fDP.jpg

remove the dot after the first h to view it

Theboybang

<snip unhelpful post>

EoghanIRL

I never had that book but here is some help .

1) average is total sum of numbers / amount of numbers or e of x/ x

So do this
1,2,3,4,5,6,7,8 +m(8) , 9 +n(9) .
Add these numbers , divide them by amount of number . Let this answer equals average .use substitution .

Hopefully this helps . I have no time for the rest.

Raspberry Fileds

11. (1+2+3+4+5+6+7+8+9+8m+9n) / (9+m+n) = 7.3

12. n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 = 1815

13. n^2 + 92 = (n+2)^2 - 100

14. Erm...

rainbowtrout

The palindromes question asks for 3 digit palindromes

So they must start and end with the same digit

There are 9 of those combinations

1_1
2_2
3_3 ... 9_9

you can put the digits 0-9 in the gap for each one [e.g. 101, 111, 121,... 191], so 10 possible digits for each of the 9 types of palindromes listed

so 9 * 10 = 90 3 digit palindromes.



Part (ii)

Palindromes between 1,000,000 and 10,000,000 will have the structure:


xyzwzyx

Again 9 choices of combinations in the X positions (as zero in the first position would be less than 10^6 and would result in a six digit number)
10 choices of numbers in the Y positions
10 choices of numbers in the Z positions
10 choices of digit for the w position in the centre

9 * 10 * 10 * 10 = 9000 palindromes between 10^6 and 10^7


Part (iii)


For each set of palindromes , i.e. all possible palindromes starting and ending with 1yzwzy1 there are 1 *10*10*10 combinations = 1000

so 1000 start and end with 1, 1000 start and end with 2 etc

So position 2125 on the list must be a palindrome starting and ending with 3.

3yzwzy3

So position 125/1000 on this list

All of the palindromes that start with y being a 0 and going through the range of digits for z and w will yield 1 * 10 * 10 combinations = 100

so that means that palindrome 125 in the set is within the second 100 of this set

so it has the combination

31zwz13

There are 10 digit options for z and 10 digit options for w, both can have the digits 0-9

10 * 10 = 100 possible combinations

The first 10 combinations would be starting with z = 0 and w being each of the digits 0-9
The second 10 combinations would start with z = 1 and 0-9 for w
The third 10 combinations would start z = 2

so with the pattern zwz for these 10 combinations you get
202
212
222
232
242 (this is the 25th on the list)

So position 2125 is

3124213

Raspberry Fileds

rainbowtrout wrote: »

The palindromes question asks for 3 digit palindromes

So they must start and end with the same digit

There are 9 of those combinations

1_1
2_2
3_3 ... 9_9

you can put the digits 0-9 in the gap for each one [e.g. 101, 111, 121,... 191], so 10 possible digits for each of the 9 types of palindromes listed

so 9 * 10 = 90 3 digit palindromes.



Part (ii)

Palindromes between 1,000,000 and 10,000,000 will have the structure:


xyzwzyx

Again 9 choices of combinations in the X positions (as zero in the first position would be less than 10^6 and would result in a six digit number)
10 choices of numbers in the Y positions
10 choices of numbers in the Z positions
10 choices of digit for the w position in the centre

9 * 10 * 10 * 10 = 9000 palindromes between 10^6 and 10^7

For part (i), as xxx is also a palindromic number, the answer is 100.

rainbowtrout

Raspberry Fileds wrote: »

For part (i), as xxx is also a palindromic number, the answer is 100.

I included that:

101, 111, 121, 131, 141, 151, 161, 171, 181, 191 = 10 palindromes for combinations starting/ending with 1.

Repeat for combinations starting 2...9.

Raspberry Fileds

rainbowtrout wrote: »

I included that:

101, 111, 121, 131, 141, 151, 161, 171, 181, 191 = 10 palindromes for combinations starting/ending with 1.

Repeat for combinations starting 2...9.

My bad...I included 0x0.

TheOwl12

Hey everyone! Sorry I didn't reply. Ye all helped a lot. He said I was right but never explained them to anyone who got them wrong... should be a problem some day but he's an alright guy for now

spurious

TheOwl12 wrote: »

Hey everyone! Sorry I didn't reply. Ye all helped a lot. He said I was right but never explained them to anyone who got them wrong... should be a problem some day but he's an alright guy for now

It will be a problem. His job is to explain to those that got it wrong where they went wrong. Does he think they will just have some sort of flash of inspiration to get it right next time?

MegLC

I'm in 6th year and I don't remember doing this type of maths? but still you should tell him you didn't get it right and you need explaining! HL maths for LC is a huge course, you need to be getting the stuff if you want to do well!