A mathematical question for you

[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.

[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?

[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.

[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 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.

[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

[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]

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

[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.

[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.