Reply to: A mathematical question for you

[MancMan]

Thanks very much to all who have helped with this one. Much appreciated. I won't have time until tomorrow to think about your replies and try to follow, but just wanted to say thanks very much.

[DeadEyedJacks]

Last question on paper

It's worth noting that this was the last question on a paper from King Edwards School, Birmingham in 2011.
11+ specimen papers

I reckon it was either put there to keep candidates busy until the allocated time was up

Or as a one mark "tie-breaker" to ensure not everyone got 100% on this section.

[motoukenin]

That is one answer. TRy http://forums.contractoruk.com/gener...ml#post2305476

Too complicated an answer, getting something into a power of 10 which all 10 year olds will know is the key here, not sure if the 10 year old that can do the transposition or get to the equation in the first place would be able to deal with life when he/she gets to 18 would be my concern so I would be pretty happy to have a 10 year old that could not solve it, in fact if I do have kids I might just give him/her this equation just to make sure they can't.

[BrilloPad]

equation is (s-9)/10 + 9x10^x = 4s

transpose to (39s+9)/90 is a power of 10

powers of 10 are 10, 100, 1000, 10000, 100000 ...etc

if s is an integer then first to solve is 100000

therefore (39s+9)/90 =100000

s = 230769

That is one answer. TRy http://forums.contractoruk.com/gener...ml#post2305476

[motoukenin]

equation is (s-9)/10 + 9x10^x = 4s

transpose to (39s+9)/90 is a power of 10

powers of 10 are 10, 100, 1000, 10000, 100000 ...etc

if s is an integer then first to solve is 100000

therefore (39s+9)/90 =100000

s = 230769

Think that's how a 10 year old is supposed to solve this, of course when he gets to 17 he will know everything anyway and should give the correct answer without any working out.

[BrilloPad]

God, when was this exam, 1890? Are you sure it was the 11 Plus and not the Cambridge Tripos?

It's quite a tricky problem. But in fact there are an infinite number of answers.

Symbolically, the problem amounts to finding integers x and n such that (with "dot" denoting multiplication) :

  9 . 10^n + x  =  4 (10 x + 9)     where  x < 10^n

Rearranging gives:

  39 x  =  9 (10^n - 4)

So since 3 || 39 (standard notation to mean 3 is the largest power of 3 dividing 39), but from the equation 3^2 | 39 x, we see that 3 | x.

So x = 3 y for some integer y and, rearranging once more, this gives :

    10^n  =  13 y + 4

The smallest positive integer n satisfying this for an integer y is n = 5, i.e. 10^5 = 13 . 7692 + 4, which presumably leads to NAT's solution.

But you can also multiply each side of the preceding equation by corresponding sides of 10^6 = 13 . 76923 + 1 any number of times, and obtain a new solution each time.

In other words, every integer solution n (i.e. for which an integer value of y is obtained) is given by n = 5 + 6 t, for t = 0, 1, 2, ... infinity

So, to be explicit, every integer satisfying the conditions of the problem is of the form 30 (10^(5 + 6 t) - 4) / 13 + 9 for t = 0, 1, 2, ...

This was my favourite answer.

[NotAllThere]

NotAllThere, please give some more detail and explanation, I want to understand this algebra

Well, the quoted solution was marginally incorrect. It should have read:

(n digits)9
x 4
9(n-1 digits)6

So the tens's digit is 6. Continue to get 230769

Sorry - can't be bothered to explain further.

[MancMan]

Please explain!

Easy

(n digits)9
x 4
9(n-1 digits)6


So the hundred's digit is 6. Continue to get 230769

NotAllThere, please give some more detail and explanation, I want to understand this algebra

[MancMan]

ChimpMaster, Please explain how you got the answer

My special number has a 9 in the units column. If I remove the 9 from the units column and place it at the left hand end of the number, but leave all the other digits unchanged, I get a new number. This new number is four times my special number. What is my special number?

--------------

So the question above is from an 11+ grammar school entry exam. I've worked out the answer but not without some pain and guesswork thrown in for good measure. I just can't imagine how my kid or any 10 year is supposed to work it out!

ChimpMaster, please explain how you got the answer

[MancMan]

I don't get it, please explain more

Hi NotAllThere and ChimpMaster

You both got the answer to this question: My special number has a 9 in the units column. If I remove the 9 from the units column and place it at the left hand end of the number, but leave all the other digits unchanged, I get a new number. This new number is four times my special number. What is my special number?

ChimpMaster got the answer, but didn't give an explanation. Could you explain how you did it please?

NotAllThere explained as below. I just don't understand the algebra, could you please explain step by step with a little more detail?


Easy

(n digits)9
x 4
9(n-1 digits)6


So the hundred's digit is 6. Continue to get 230769

[vetran]

if you're prepared to pay for them to go to a £12k/year school it makes sense to spend a few hundred ££ on making sure they get in, no?

yep.

It was just that they start at 6 to get into grammar. It does seem to be predominantly Indian families that do that.

[pr1]

That's what happens round here.

if you're prepared to pay for them to go to a £12k/year school it makes sense to spend a few hundred ££ on making sure they get in, no?

[vetran]

If it was true to the original 11+ there would have been no preparation. I would guess these days it's possible to pay a tutor to prep your 6 YO on past papers.

That's what happens round here.

[BrilloPad]

I sometimes get roped in to do recruitment for client co.
I will be using this question - its not knowledge specific and not too difficult.

Have you interviewed anyone and had them accept the job?

Do you really want anyone who would work with an ubercretin?

[sasguru]

I sometimes get roped in to do recruitment for client co.
I will be using this question - its not knowledge specific and not too difficult.