Haberdasher's 2014 Maths Q: adding sequences of numbers
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Haberdasher's 2014 Maths Q: adding sequences of numbers
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)
Many thanks in advance
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)
Many thanks in advance
Re: Haberdasher's 2014 Maths Question
S(1,3) = 1+ 2 + 3
Then follow their instructions.
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 ...
Then follow their instructions.
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 ...
Re: Haberdasher's 2014 Maths Question
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.
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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Re: Haberdasher's 2014 Maths Question
I think the last question is quite elegant.sadiamek wrote: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)
Many thanks in advance
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 9:11 pm, edited 1 time in total.
Re: Haberdasher's 2014 Maths Question
I deliberately didn't give an answer because it's better to help by giving a hint!
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Re: Haberdasher's 2014 Maths Question
Edited my post after Guest55's suggestion
Re: Haberdasher's 2014 Maths Question
sadiamek wrote: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)
Many thanks in advance
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
Re: Haberdasher's 2014 Maths Question
didn't read the question! Thanks
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Re: Haberdasher's 2014 Maths Question
Thanks Elibet