Challenge Question 1 (2 Viewers)

[acmilan]

Here is the first question, dont really know if you'll find it easy or not as i dont know all your standards:

A person goes to withdraw a dollars and b cents, but by mistake receives b dollars and a cents. After spending 5 cents, they realise they have twice as much as they asked for. How much did they initially want?

(Hint: a and b must be integers and 0 <=a <100 and 0 <= b < 100)

Please PM with any answer you get as we dont want to ruin it for people that dont get it as fast. Also note that the answer contains nothing that an ordinary 2 unit student wouldnt understand, just some lateral thinking.

EDIT: Also dont just sub in random numbers until you get a figure that matches the description, there has to be mathematical proof that your answer is correct

 

[Slidey]

Meh I can't get an answer without pronumerals in it.

 

[acmilan]

Sorry i made a type, it should have been he received b dollars and a cents, not y cents

 

[dawso]

how did i know that was comin.....*sighs*..... disregard my pm m8

try again dawso.....

btw, i shud get some credit 4 all the work i did on the old question, lol, its only affected me and steele, pffft

 

[dawso]

damn... here I am subbing in different numbers...

I should get back to my uni stuff.....

lol, u keep out of this, year 12s only!!!

 

[withoutaface]

Nice question Vince, I can't see an immediate solution but I remember a helpful method for calculating things such as ABCD and comparing them to DCBA from when I was doing olympiad questions. But yeah, back to my Linear Algebra study. Have fun guys

EDIT: http://www.angelfire.com/ab7/fourunit/imo-all.pdf
1978 Q A1 should be doable by a capable 4u student

 

[dawso]

lol, no, not a 'nice question' lol, a bloody annoying one, i got it down 2 one simultaneous equation but cant work out how 2 get another and geez, i gotta go do other work not think about this question, someone come up with a solution quickly please!!!

 

[velox]

lol, u keep out of this, year 12s only!!!

keep out, no 4u guys! lol

 

[nick1048]

ahhhhhhh freaking simultaneous equations... I can't seem to get anywhere through them no matter what I do. Trial and error... dammit I'm so tempted to use that... bah I give.

 

[acmilan]

No takers?

If not ill post answer tonight

 

[Slidey]

A person goes to withdraw a dollars and b cents, but by mistake receives b dollars and a cents. After spending 5 cents, they realise they have twice as much as they asked for. How much did they initially want?

"(Hint: a and b must be integers and 0 <=a <100 and 0 <= b < 100)"

Form an equation:
100b+a-5=a200+2b
98b-199a=5
b=(5+199a)/98
Arithmetic series.
First term=c=204/98, d=199/98
T_n=(199n+5)/98
However, we do not know what the nth term is - we don't know what to look for.
But we do know that it is a natural number, so the decimal part shall be 0.

Make an arithmetic series from the part of each term thatis less than 1, then find when T_n=1:
(1) n=1, T_1=204/98
(2) a=2, T_2=403/98
Common difference, d, is equal to the decimal part of (2)-(1), which is 3/98
First term was when a=1 so 204/98, but we want the decimal part of this: 4/49 thus the arithmetic series we want is:
T_n=4/49 + (n-1)*(3/98)
We shall solve this for T_n=1 since that value of n will be when we get the first natural number (and hence will be our solution). Remembering that T_n=b and n=a.
T_n=4/49 + (n-1)*(3/98)=1
(n-1)*(3/98)=45/49
n-1=2*45/3
n=31 #
Substitute into original equation, b=(5+199a)/98:
a=31, b=63

So the person wanted: $31.63
Let's test that:
2*(31.63)+0.05=63.31, so it works.

EDIT: Cleaned up my solution.

 

[Trebla]

SORRY!!!!!!
(BRAVO!) lol

 

[acmilan]

Congrats to Slide_Rule, he used a different method to me, but it still gave correct result.

 

[Trebla]

...and the message with the solution was deleted...

 

[Slidey]

I'll repost it if you want. I thought I'd let others have a go but that doesn't look like it's going to happen. I thought perhaps they'd be able to solve it the way acmilan intended. I'd be interested to see his solution.

 

[acmilan]

My answer was just basically an algebra bash. I also undeleted Slide_Rule's answer

 

[Slidey]

Wow. That's nifty, acmilan (algebra bash or not, it's an interesting solution). I'll have to take a thorough look at it later.

I've cleaned up my solution. No more ugly decimals.

 

[acmilan]

I have unofficially introduced you to Diophante Equations

 

[withoutaface]

Rep points for the first to solve my question

 

[withoutaface]

I have unofficially introduced you to Diophante Equations

What a horrible thing to do!