A mathematical question for you

[NotAllThere]

And they have to do that without a calculator? Jeepers.

Simple multiplication. I did it on a piece of paper. There's no guesswork. You know the last digit of the initial number is 9. Any number ending in 9, multiplied by 4 will yield a number ending in 6. Now you have the second digit.

If you construct the notional multiplication it becomes obvious.

[ChimpMaster]

Simple multiplication. I did it on a piece of paper. There's no guesswork. You know the last digit of the initial number is 9. Any number ending in 9, multiplied by 4 will yield a number ending in 6. Now you have the second digit.

If you construct the notional multiplication it becomes obvious.

Thanks, I'll tell my 10 year old to do that

[NotAllThere]

"Obvious" is a mathematical term meaning "it's obvious to me". Similarly "The proof is an exercise left to the reader" means "I'm sure it's true but can't be arsed to prove it".

[BrilloPad]

"Obvious" is a mathematical term meaning "it's obvious to me". Similarly "The proof is an exercise left to the reader" means "I'm sure it's true but can't be arsed to prove it".

I got that from my A level teacher. Try to get a formula for circumference of an ellipse. It vexed me for 2 days.....

[NotAllThere]

Thanks, I'll tell my 10 year old to do that

To be fair - if he can understand how the answer is figured out, he's doing well. I guess this question separates out those who understand multiplication (to an extent) from those who can simply mechanically apply it.

[OwlHoot]

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?

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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!

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, ...

[NotAllThere]

My son has just pointed out 230769230769230769 is a solution. But your general solution is worthy of respect.

[ctdctd]

My son has just pointed out 230769230769230769 is a solution. But your general solution is worthy of respect.

Now PM that number to Brillo as a target for the Counting thread - that should keep him going for a while.

[Old Greg]

Now PM that number to Brillo as a target for the Counting thread - that should keep him going for a while.

I've already deleted some entries.

[GJABS]

Now PM that number to Brillo as a target for the Counting thread - that should keep him going for a while.

230769230769230770