Maths Homework Stumper

[SpeckledJim]

I know I’m going to kick myself, but can anyone see the rule please?




Using a spreadsheet we’ve found one combination that sums to 100. How many are there?

[anonymous-user]

I’ve never been a fan of doing other peoples homework for them

[saaby93]

Some help here
http://www.robertmellors.notts.sch.uk/wp-content/u...

if the squares are labelled

a b
c d

b+d+c+d= 10
try d = 2, b= 1, c= 5

then
a1 +52 +a5+12 = 100
a1+a5= 36
ah
try again

Edited by saaby93 on Saturday 11th June 11:11

[Narcisus]

If I wasn’t st at maths I would do it for you without any of the sanctimonious bullst :-)

[anonymous-user]

Narcisus said:

If I wasn’t st at maths I would do it for you without any of the sanctimonious bullst :-)

Shhh.... that might be why I posted what I did.

[CobolMan]

Call the boxes
a b
c d

You're looking to solve
(10a + b) + (10c + d) + (10a + c) + (10b + d) = 100

Which simplifies to 20a + 11(b + c) + 2d = 100

20a is always going to be even
2d is also always going to be even
11(b +c) is always going to be odd

Even + even + odd will always be odd so it can't be done.

[TheInternet]

CobolMan said:

11(b +c) is always going to be odd

I think two solutions.

[Ultuous]

I don't know how maths is 'marked' these days, but particularly given the age range, I'd go with logic rather than strict algebra...

a b
c d

What needs to be totaled is:

ab
cd
ac
bd

We're not interested in the '10s', as it's only b+d+c+d part that's of interest to whether the result is odd or even (the 10's will just add 10's to that)...

So b+d+c+d or b + c + 2d

2d will always be even (1 x 2 = 2, 2 x 2 = 4, 3 x 2 =6 etc.), so that can be discounted.

That leaves us with just b and c

odd = odd is always even (e.g. 1+1, 1+3, 3+5)
odd + even is always odd (1+2, 3+4 etc.)
even + even is always even

So the 'quick way' is that if either if squares b and c contain odd and even numbers, the result is odd, otherwise the result is even.

Edited by Ultuous on Saturday 11th June 11:48

[TwigtheWonderkid]

14
27

14+27+12+47 = 100

I can't come up with any others.

Edited by TwigtheWonderkid on Saturday 11th June 12:04

[Sporky]

Recovering ground others have mentioned.

Total is 20a + 11(b+c) +2d

So if just one of b & c is odd, the answer is odd.

For the total of 100 you start with some "obvious" (horrible word, sorry) thinking:

a can only be 1, 2, 3, or 4, because if a was 5 the total would be over 100 straight away.

If a=3 then b and c must be 2 and 4 for the answer to be odd, but that's a total of 126 already, so a can't be 3.

If a=2 then try b and c are 1 and 3, adding up to 84, leaving d=8. That's two solutions because it doesn't matter which way around b and c are. Clearly there's no other solution for a=2 as the next options for b and c put the total over 100.

If a=1 then try b and c are 2 and 4, giving 86 this far, and d=7. Two more solutions, and clearly no more.

Four solutions. Gold star for Sporky.

[CobolMan]

TheInternet said:

I think two solutions.

Oh bo!!ocks, I didn't think that one through properly

[saaby93]

Challenge for 7-11 yr old
whats the collective age rating here to crown sporky?

[Sporky]

saaby93 said:

Challenge for 7-11 yr old
whats the collective age rating here to crown sporky?

Full disclosure; I am a little bit older than -4.

[Narcisus]

garyhun said:

Narcisus said:

If I wasn’t st at maths I would do it for you without any of the sanctimonious bullst :-)

Shhh.... that might be why I posted what I did.

I have the greatest respect for anyone who can make sense of a jumble of letters and numbers