Please help me with DSs maths homework (Y5)

(33 Posts)
LynetteScavo Thu 20-Nov-08 15:59:45

Choose four numbers from the following;

1 2 3 4 5 6 7 8 9

put each digit in a 2x2 box.

This makes two 2-digit numbers reading across, and two 2-digit numbers reading down.

Add up all four of the numbers.

(for example choosing 1 2 4 & 7 12 + 47 + 14 + 27 = 100)

How many differnt ways of making 200 can you find?

(Please help, I'm feeling poorly, my head hurts just reading this homework and DH is comming home late.sad)

OP’s posts: |
cat64 Thu 20-Nov-08 16:04:33

Message withdrawn

bellavita Thu 20-Nov-08 16:05:48

If no one can help, my DS just gone into Yr7 will be home shortly. I will ask him ('cos I have no idea how to do it} blush

Flier Thu 20-Nov-08 16:07:31

if you do it on a spreadsheet it may make it easier - if this is allowed

stealthsquiggle Thu 20-Nov-08 16:07:47

I take it DS is struggling?

I would have to sit down and play and see if you can derive a rule, which is presumably the point.

<<waits for maths teacher to magically appear and explain how to reverse-engineer it>>

Take a paracetamol and coffee break and then try again.

LynetteScavo Thu 20-Nov-08 16:08:58

Thankyou bellavita. smile

OP’s posts: |
Flier Thu 20-Nov-08 16:13:22

bloomin' eck! can't get any figs to add to 200!

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bellavita Thu 20-Nov-08 16:23:43

The first thing to do is to eliminate which number cannot go in the bottom right square (it can't be 9 for instance)

If you label the numbers along the top a & b and along the bottom c & d the sum is always ab + cd + ac + bd. In order for the total to be 200 b + c + d + d must be a multiple of 10.

Does this make sense?

witchandchips Thu 20-Nov-08 16:26:58

forget about the box, need 4 numbers that add up to 200, each of which have to be between 10 and 99 (inclusive)

don't think there is any rule, think the point of the exercise is to get your dc to balance the sum, so if he takes a number smaller he has to make another number bigger iyswim

so 10 11 90 89
make the first number bigger (say 10 to 12) then have to substract 2 from the other numbers

witchandchips Thu 20-Nov-08 16:34:05

forget my last post did not read the problem properly

but its not ab etc is 10a+b etc isn't it?

LynetteScavo Thu 20-Nov-08 16:37:14

bellavita - we were with you utnitl the last sentence.

OP’s posts: |
bellavita Thu 20-Nov-08 16:39:28

blush - got to admit DS did not have a clue, so I msn'd my husband - physics/chemistry/maths is his thing and he gave the formula.

I did not understand any of it blush

witchandchips Thu 20-Nov-08 16:43:17

a has to be less than 5
b+c+d has to equal either 10 or 20

its hard

Flier Thu 20-Nov-08 16:45:31

bellavita,

is this part correct? "In order for the total to be 200 b + c + d + d must be a multiple of 10"

b+c+d+d must be a multiple of 10?

LynetteScavo Thu 20-Nov-08 16:51:04

You can't use 0, so there are no multiples of 10, are there? hmm

I have emailed my cousin who studied maths at Cambridge.

OP’s posts: |
Flier Thu 20-Nov-08 16:51:05

I have 2 so far

bellavita Thu 20-Nov-08 16:51:17

Cannot answer as DH is now on is way home from work. I thought I had copied it word for word from msn (could not copy and paste it for some reason) and I do not have the convo anymore.

DH will probably be home around 5.30 - will ask when he comes in.

Flier Thu 20-Nov-08 16:53:16

3

Flier Thu 20-Nov-08 16:54:51

I have 4 now, and top left number is 2, all number when added together add to 19.

witchandchips Thu 20-Nov-08 16:56:10

following bellavita labelling we have

a b
c d

which makes (10a+b)+(10b+d)+(10a+c)+(10b+d)
= 20a+11b+11c+d

20a has to end in zero to 11b+11c+d must end in zero as well. 11b ends with b (11, 22, 33 etc) so this means b+c+d must end in zero

Flier Thu 20-Nov-08 17:03:13

not quite, bella, the 4 squares I have got to add to 200, b+c+d=17. a+b+c+d=19.

Flier Thu 20-Nov-08 17:13:12

sorry, that was to witchandchips

bellavita Thu 20-Nov-08 17:59:35

That's correct Flier, all solutions have a+b+c+d=19
d=a+1.
b+c+(2d)=20
(2
a)+b+c=18

So
a=1, b=9, c=7, d=2
a=2, b=8, c=6, d=3
a=2, b=9, c=5, d=3
a=3, b=7, c=5, d=4
a=4, b=7, c=3, d=5
a=4, b=8, c=2, d=5
a=4, b=9, c=1, d=5
a=5, b=7, c=1, d=6
a=6, b=4, c=2, d=7
a=6, b=5, c=1, d=7
a=7, b=1, c=3, d=8

all work

Bellavita's other/better half

stealthsquiggle Thu 20-Nov-08 17:59:37

Granted I am just playing with a spreadsheet rather than trying to derive the formula very scientifically, but I cannot get any answers where all 4 numbers are different angry

stealthsquiggle Thu 20-Nov-08 18:00:28

damn x-post. Scientific approach wins grin

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