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 Posted: Wed Nov 26, 2014 10:11 am

Joined: Wed Nov 28, 2012 5:39 pm
Posts: 8
Can anyone help please. Difficult to explain to my son.

We write S(2,5) as an abbreviation for 2+3+4+5 so that S(2,5) = 14.

Similarly, S(6,39) = 6+7+8+9+......+38 +39 = 765

Work out:

S(1,3)

S(6,40)

S(7,38)

S(1,2) - S(2,3) + S(3,4) - S(4,5) + ....... - S(18,19) + S(19,20)

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 Post subject: Re: Haberdasher's 2014 Maths QuestionPosted: Wed Nov 26, 2014 10:23 am

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 14010
S(1,3) = 1+ 2 + 3

For the last one, start writing it out and you wll see a lot of the terms 'disappear'.

S(1,2) - S(2,3) + S(3,4) - S(4,5) + ....... - S(18,19) + S(19,20) = 1 + 2 - (2 + 3) + (3+ 4) .... etc

so which terms remain ...

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 Post subject: Re: Haberdasher's 2014 Maths QuestionPosted: Wed Nov 26, 2014 7:10 pm

Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3826
Add in pairs so that you don't have to do lots of tedious addition
S(6,40)
6+40=46
7+39=46
8+38=46
...
22+24=46
17 x 46 + 23 (the odd one left)

S(7,38)
7+38=45
8+37=45
...
22+23=45
16 x 45 =

Much less tedious than adding and takes significantly less time.

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 Post subject: Re: Haberdasher's 2014 Maths QuestionPosted: Wed Nov 26, 2014 7:46 pm

Joined: Wed Jan 25, 2012 7:43 pm
Posts: 204
Can anyone help please. Difficult to explain to my son.

We write S(2,5) as an abbreviation for 2+3+4+5 so that S(2,5) = 14.

Similarly, S(6,39) = 6+7+8+9+......+38 +39 = 765

Work out:

S(1,3)

S(6,40)

S(7,38)

S(1,2) - S(2,3) + S(3,4) - S(4,5) + ....... - S(18,19) + S(19,20)

I think the last question is quite elegant.

S(1,2) - S(2,3) = 1 - 3
S(1,2) - S(2,3) + S(3,4) = 1+ 4
S(1,2) - S(2,3) + S(3,4) - S(4,5) = 1 - 5

Expanding similarly
S(1,2) - S(2,3) + S(3,4) ............ + S(19,20) = 1 + ?

Question like S(7,38) is quite brutal for a primary school child IMO.

Method suggested by moved of course works. To write down / count the number of pairs may not be that easy.

A secondary school kid could have tried to solved it using the formula for sum of first n natural numbers I,e, n(n+1)/2

S(7,38) therefore would have been 38*39/2 - 6*7/2. How a primary school child is supposed to know this though .

Last edited by ConcernedDad on Wed Nov 26, 2014 8:11 pm, edited 1 time in total.

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 Post subject: Re: Haberdasher's 2014 Maths QuestionPosted: Wed Nov 26, 2014 7:59 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 14010
I deliberately didn't give an answer because it's better to help by giving a hint!

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 Post subject: Re: Haberdasher's 2014 Maths QuestionPosted: Wed Nov 26, 2014 8:13 pm

Joined: Wed Jan 25, 2012 7:43 pm
Posts: 204
Edited my post after Guest55's suggestion

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 Post subject: Re: Haberdasher's 2014 Maths QuestionPosted: Wed Nov 26, 2014 8:31 pm

Joined: Mon Mar 03, 2014 5:50 pm
Posts: 155
Can anyone help please. Difficult to explain to my son.

We write S(2,5) as an abbreviation for 2+3+4+5 so that S(2,5) = 14.

Similarly, S(6,39) = 6+7+8+9+......+38 +39 = 765

Work out:

S(1,3)

S(6,40)

S(7,38)

S(1,2) - S(2,3) + S(3,4) - S(4,5) + ....... - S(18,19) + S(19,20)

They have already told us s(6,39) is 765. Therefore s(6,40) is simply 765+40
Similarly s(7,38) is 765-6-39

The last one is easy 1+20

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 Post subject: Re: Haberdasher's 2014 Maths QuestionPosted: Wed Nov 26, 2014 9:39 pm

Joined: Fri Oct 12, 2007 12:42 pm
Posts: 3826

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 Post subject: Re: Haberdasher's 2014 Maths QuestionPosted: Thu Nov 27, 2014 3:30 am

Joined: Wed Jan 25, 2012 7:43 pm
Posts: 204
Thanks Elibet

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