HSC Mathematics Marathon (5 Viewers)

hup · Sep 28, 2011

artosis said:
haha good question. Baulko 2011 trial?
took me quite awhile to get it out

yea and there is a very simple way to do it

AAEldar · Sep 28, 2011

hup said:
$\\ Prove\ by\ induction\ that\\ 7^n+3n(7^n)-1\ is\ divisible\ by\ 9\ for\ n\geq 1$

$\text{Prove by induction that }7^n+3n(7^n)-1\text{ is divisible by 9 for }n\geq1 \\ \text{Assume true for }n=k \\ \text{i.e. }7^k+3k(7^k)-1=9p\text{ where p is an integer} \\ 7^k(1+3k)-1\;\Rightarrow\;7^k=\frac{9p+1}{1+3k} \\ \text{Prove true for }n=k+1 \\ \text{i.e. }7^{k+1}(4+3k)-1 \\ =\frac{7(9p+1)(4+3k)}{1+3k}-1\;\;\;\;\;\text{from above} \\\\ =\frac{7(36p+27pk+4+3k)}{1+3k}-1 \\\\ =\frac{252p+189pk+28+21k}{1+3k}-\frac{1+3k}{1+3k} \\\\ =\frac{252p+189pk+18k+27}{1+3k} \\\\ =\frac{9(28p+21pk+2k+3)}{1+3k}\;\;\;\;\;\;\;\text{which is divisible by 9}$

Left out proving for n=1 and the other statements. That the simple way?

apollo1 · Sep 28, 2011

hup said:
$\\ Prove\ by\ induction\ that\\ 7^n+3n(7^n)-1\ is\ divisible\ by\ 9\ for\ n\geq 1$

hey i did this question when i sat the baulko trial i was so happy to get it out.

t0nyy · Sep 28, 2011

AAEldar said:
$\text{Prove by induction that }7^n+3n(7^n)-1\text{ is divisible by 9 for }n\geq1 \\ \text{Assume true for }n=k \\ \text{i.e. }7^k+3k(7^k)-1=9p\text{ where p is an integer} \\ 7^k(1+3k)-1\;\Rightarrow\;7^k=\frac{9p+1}{1+3k} \\ \text{Prove true for }n=k+1 \\ \text{i.e. }7^{k+1}(4+3k)-1 \\ =\frac{7(9p+1)(4+3k)}{1+3k}-1\;\;\;\;\;\text{from above} \\\\ =\frac{7(36p+27pk+4+3k)}{1+3k}-1 \\\\ =\frac{252p+189pk+28+21k}{1+3k}-\frac{1+3k}{1+3k} \\\\ =\frac{252p+189pk+18k+27}{1+3k} \\\\ =\frac{9(28p+21pk+2k+3)}{1+3k}\;\;\;\;\;\;\;\text{which is divisible by 9}$

Left out proving for n=1 and the other statements. That the simple way?

definitely not lol
also idk if that can be considered right because you have no way of knowing that huge fraction is always a whole number

apollo1 said:
hey i did this question when i sat the baulko trial i was so happy to get it out.

lets see ur working

K4M1N3 · Sep 28, 2011

Just going to skip to the second step:
2. Let $7^k(1+3k)-1=9M$ Where M is an integer

Prove true for n = k+1

$7^{k+1}(1+3(k+1))-1$

$7.7^k(4+3k)-1$

$7(7^k(1+3k)-1+1+3.7^k)-1$

Using result from 2.

$7(9M+1+3.7^k)-1$

But since $7^k(1+3k)-1=9M$

$7^k = \frac{9M+1}{1+3k}$

Now subbing back in

$7(9M+3.\frac{9M+1}{1+3k})+6$

$7(9M)+ \frac{21(9M)+21}{1+3k}+6$

$7(9M)+ \frac{21(9M)+21+6+18k}{1+3k}$

$9(7M+ \frac{21M+3+2k}{1+3k})$

therefore divisible by 9.

AAEldar · Sep 28, 2011

t0nyy said:
definitely not lol
also idk if that can be considered right because you have no way of knowing that huge fraction is always a whole number

lets see ur working

Yea I wasn't sure about that fraction, but I cbf doing it again >.>

apollo1 · Sep 28, 2011

let me find it. its under a stack of papers i did. i will post my solution in 20 minutes after dinner. tOnyy do u go to baulko?

t0nyy · Sep 28, 2011

apollo1 said:
let me find it. its under a stack of papers i did. i will post my solution in 20 minutes after dinner. tOnyy do u go to baulko?

nah i just know the working cause i asked hup for solution

apollo1 · Sep 28, 2011

posting my solution in a sec. im using latex.

apollo1 · Sep 28, 2011

from assumption:
$7^{k}+3k(7^{k})-1 = 9A$
prove true for k+1:
hence prove $7^{k+1}+3(k+1)(7^{k+1})-1$ is divisible by 9
$7^{k+1}+3(k+1)(7^{k+1})-1$
$7.7^{k}+3k.7^{k+1}+3.7^{k+1}-1$
$28.7^{k}+21k.7^{k}-1$
$7(7^{k}+3k.7^{k}-1)+21.7^{k}+6$
$7(9A)+21.7^{k}+6$
$7(9A)+21.(7^{k})+6$
$7(9A)+21.(9A-(3k.7^{k}-1))+6$
$9(28A-7k.7^{k}+3)$

hence divisible by 9 (since A and k integers).

conclusion: blah blah blah

hup · Sep 28, 2011

apollo1 said:
from assumption:
$7^{k}+3k(7^{k})-1 = 9A$
prove true for k+1:
hence prove $7^{k+1}+3(k+1)(7^{k+1})-1$ is divisible by 9
$7^{k+1}+3(k+1)(7^{k+1})-1$
$7.7^{k}+3k.7^{k+1}+3.7^{k+1}-1$
$28.7^{k}+21k.7^{k}-1$
$7(7^{k}+3k.7^{k}-1)+21.7^{k}+6$
$7(9A)+21.7^{k}+6$
$7(9A)+21.(7^{k})+6$
$7(9A)+21.(9A-(3k.7^{k}-1))+6$
$9(28A-7k.7^{k}+3)$

hence divisible by 9 (since A and k integers).

conclusion: blah blah blah

that is better and you can shorten it

$\\assume\\\\ 7^k(1+3k)-1=9M \\\\for\ n = k+1\\\\ 7^{k+1}+3(k+1)(7^{k+1})-1\\\\ =7^{k+1}(3k+4)-1\\ =7^k(21k+28)-1\\ =7^k[(1+3k)+(27+18k)]-1\\ =7^k(1+3k)-1+7^k(9)(3+2k)\\ =9M+9(7^k)(3+2k)\\\\ which\ is\ divisible\ by\ 9$

apollo1 · Sep 28, 2011

hup said:
that is better and you can shorten it

$\\assume\\\\ 7^k(1+3k)-1=9M \\\\for\ n = k+1\\\\ 7^{k+1}+3(k+1)(7^{k+1})-1\\\\ =7^{k+1}(3k+4)-1\\ =7^k(21k+28)-1\\ =7^k[(1+3k)+(27+18k)]-1\\ =7^k(1+3k)-1+7^k(9)(3+2k)\\ =9M+9(7^k)(3+2k)\\\\ which\ is\ divisible\ by\ 9$

cool. hold on ill find a question to poast

artosis · Sep 28, 2011

From the Girraween 2011 Trial (No solutions) =[

In how many ways can the letters of the word MMAATTHH be arranged
(i) Without restriction
(ii) With at least one pair of repeated letters together
(iii) With no pair of repeated letters together

Can anyone confirm my answers:
(i) 2520
(ii) 396
(iii)864

largarithmic · Sep 28, 2011

I think that question is actually reasonably hard, I can't think of a way to do it without using the "inclusion-exclusion" principle, something that isn't really in the course (although we did it in class in case it might come up in a problem like that).

Essentially think of a venn diagram like thing with 4 circles (although it doesnt quite work that simply) with each circle representing say, the cases where the letters M and M are put together. Then take their intersections, etc.

Basically:
Number of things where the X and X are together for a particular X is 7!/2!2!2!.
Number of things with XX and YY are 6!/2!2!
Number of things with XX and YY and ZZ together are 5!/2!
Number of things with XX, YY, ZZ, WW all together is 4!

Then the number of things with at least one pair of letters together is
4 * 7!/2!2!2! - 6 * 6!/2!2! + 4 * 5!/2! - 4! = 1656 (inclusion exclusion principle?)

And then when there are no repeated letters together is 2520 - 1656 = 864

apollo1 · Sep 28, 2011

largarithmic said:
I think that question is actually reasonably hard, I can't think of a way to do it without using the "inclusion-exclusion" principle, something that isn't really in the course (although we did it in class in case it might come up in a problem like that).

Essentially think of a venn diagram like thing with 4 circles (although it doesnt quite work that simply) with each circle representing say, the cases where the letters M and M are put together. Then take their intersections, etc.

Basically:
Number of things where the X and X are together for a particular X is 7!/2!2!2!.
Number of things with XX and YY are 6!/2!2!
Number of things with XX and YY and ZZ together are 5!/2!
Number of things with XX, YY, ZZ, WW all together is 4!

Then the number of things with at least one pair of letters together is
4 * 7!/2!2!2! - 6 * 6!/2!2! + 4 * 5!/2! - 4! = 1656 (inclusion exclusion principle?)

And then when there are no repeated letters together is 2520 - 1656 = 864

this question is slightly annoying bcuz part ii is hard and if you get it wrong and use complementary to get part iii you end up getting both wrong.

apollo1 · Sep 28, 2011

largarithmic are u an olympiad student?

Drongoski · Sep 28, 2011

apollo1 said:
largarithmic are u an olympiad student?

Must be!

apollo1 · Sep 28, 2011

Drongoski said:
Must be!

thats what im thinking. his responses are just way too confident for a normal hsc student.

hup · Sep 29, 2011

apollo1 · Sep 29, 2011

hup said:
$\\Determine\ the\ min\ and\ max\ values\ of\ f(x)=3x^4-20x^3+36^2\\\ and\ sketch\ f(x)\ for\ -1\leq x\leq 4\\\\ Obtain\ the\ condition\ that\ f(x)=k\ has\ four\ distinct\ roots.\\ Prove\ that\ when\ this\ condition\ is\ satisfied;\\ the\ difference\ between\ the\ greatest\ and\ least\ of\ these\ four\ roots\ is\ less\ than\ \frac{4}{3}\sqrt{10}.$

is that supposed to be a 36x^2?

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