A number of things

Posted: 29 October 2011 in personal, skills
Tags: , ,

Skip this if you’re not fascinated by numbers.

I’m serious – this subject has been known to have a more soporific effect than laudanum.

OK, still with me?

I was talking with (isn’t that a great way of being polite, instead of saying “I was having a small argument with…”?) a friend last night about something and as part of it, said I’d prove that 1 = 2, and proceeded to so so, but more about that in a moment.

But it started me thinking about mathematics and the tricks those of who have a grasp of it use. Of course, I use mathematics and arithmetic synonymously, and they’re not the same, at all. But forgive me for using them this way, just this once? Thanks.

Now, I had a great maths teacher at school. He understood that to get kids interested in maths as a subject, he had to make it interesting as a subject. And to that end, he taught his class what he called the ‘tricks of the trade’.

So, for example, he taught us how to discover whether any number was divisible by any number between 2 and 12.

It remains a mystery to me how everyone doesn’t know this, but:

2: duh, the number’s even

3: if the number’s digits sum to a number divisible by 3, the number itself is divisible by 3.

4: If the last two digits of the number are divisible by 4, the whole number is.

5: the number ends in 5 or 0.

6: if the number’s even and the number’s digits sum to a number divisible by 3, the number itself is divisible by 6.

7: hmm, he didn’t teach us this one.

8: If the last three digits of the number are divisible by 8, the whole number is.

9: if the number’s digits sum to a number divisible by 9, the number itself is divisible by 9.

10: the number ends in a 0.

11. If the sum of every other digit, starting with the first, is either equal to the sum of every other digit starting with the second, or the difference is exactly divisible by 11, then the number is evenly divisible by 11. Try 13,057. 1+0+7 = 3+5, therefore it should divide evenly by 11. And indeed it does: 13,057 ÷ 11 = 1,187. Take 92,807. (9+8+7) – (2+0) = 22, therefore it should divide evenly by 11. And it does: 92,807 ÷ 11 = 8,437.

12: if the number’s digits sum to a number divisible by 3, and the last two digits of the number are divisible by 4, the whole number is exactly divisible by 12.

Except… he didn’t teach us how to quickly find if a number was divisible by seven. I even told Philip, when teaching him those same tricks, that there wasn’t a quick way.

Well, I was wrong, and it took me until the age of 40 to discover an incredibly easy way… I have no idea why it works, but it does.

Take the number’s final digit and double it. Subtract that from the rest of the digits and if you end up with a number divisible by 7, you’re home and dry.

Take the number 364. Double the final digit and you get 8. Subtract that from the first two digits: 36 – 8 = 28. And what do you know? 28 is divisible by 7, so 364 is exactly divisible by 7.

903? 90 minus 6 (3 doubled) = 84, so 903 is divisible by 7.

Look, I told you it was boring; don’t say I didn’t warn you. Be grateful, I could have taught you a quick way of working out the two-digit cube root of any number between 1,000 and 970,299.

Anyway, that proof.

I have no doubt that some of you will spot the flaw in this fairly quickly, but it’s genuinely astonishing to me how many people don’t.

(1) Let a = b

(2) Multiply both sides by b… ab = b²

(3) Subtract a² from both sides… ab – a² = b² – a²

(4) Factorise… a (b – a) = (b + a)(b – a)

(5) Divide both sides by (b – a)… a = b + a

(6) Since a = b… a = 2a

(7) Divide by a… 1 = 2

Ta-da!

Yeah – I know, but be fair, I said there was a flaw…

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Comments
  1. Littlepurplegoth says:

    (cuts and pastes whole post into notes knowing it will be useful at some point)

  2. Kate the Short says:

    But, you can’t divide by zero, can you?

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