h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

I've got a test coming up, and it's been years since I was in school. So, I seem to have forgotten some basic mathematical rules.

I'm not asking for the answer to this question, just the steps I need to take to get the answer.

Here's the equation: ( (10 = 5) (10 + 5) )

Do I first add the numbers in the inner sets of parentheses, then multiply their sums?

And of course, I've typed it incorrectly!

It's ( (10 + 5) (10 + 5) )

Thanks!

I don't know how a mathematician di the saem as a computer programmer would the syntax it would be

(10 + 5) * (10+5) = 255
10 + 5 = 15
15 * 15 = 225

Without the Parentheses:

10 + 5 * 10 + 5

10 + 5 = 15
15 * 10 = 150
150 + 5 - 155

The calculation inside the parentheses are done first. The external parentheses are superfluous. You would only need the external parentheses if the formula had another calculation:

((10 + 5) * (10+5)) / 5

then the answer would be 45 (255/5)

Without the parentheses it would be 31 (155/5)

Donald

Actually, without the parentheses the computer would always evaluate the multiplication first. So I think you'd get

10 + 5 * 10 + 5 = 10 + 50 + 5 = 65

without the parentheses, although maybe that depends on the computer you're using. But with the parentheses, you are correct.

'Do I first add the numbers in the inner sets of parentheses, then multiply their sums? '

yep, trust me I was a fantastic mathematician at school, before the virus got me!

Keep it simple, each bracket first, then deal with the results.

BODMAS isn't it?
Brackets <?> Division, Multiplication, Addition, Subtraction

I think the O is 'of' but then that misses out exponents. but for the simple 4 plus parenthesis, BODMAS works fine.

Of course, you could rewrite it as

10 5 + 10 5 + *

just to be absolutely clear.

What I learned was BIDMAS - 'I' being indices, the small numbers *over* the main ones - eg the "2" in E = mc2- so the O in yours stands for "over" I think.

Without thinking like a computer, the way is to find the sum of all the possible combinations of multiplying a number in the first bracket by a number in the second bracket.

So (a + b)(c + d)

= ac + ad + bc + bd

This works for any number of terms in the brackets.

B