A strategy for the following Maths Question
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A strategy for the following Maths Question
The last question in my daughters very last exam after three months was the following:
Add together every number between one and one hundred including one and one hundred.
So of course she came out expecting me to have a strategy ready for answering this.
I suggested doing the number bonds first, 99 and 1, 98 and 2 etc, anyone suggest a quicker method? She had answered every question on every paper for three months so was gutted to get this at the end. She did guess but with no real prospect of getting it correct.
Add together every number between one and one hundred including one and one hundred.
So of course she came out expecting me to have a strategy ready for answering this.
I suggested doing the number bonds first, 99 and 1, 98 and 2 etc, anyone suggest a quicker method? She had answered every question on every paper for three months so was gutted to get this at the end. She did guess but with no real prospect of getting it correct.
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The method is to add up the opposite ends...
1,2,...., n
1+n, 2+(n-1), 3+(n-2), .... etc.. Each of these pairs adds up to 1+n
There are n/2 such pairs....
Sum is therefore... n(n+1)/2
Here... it's even easier as you know n=100
1+100, 2+99, 3+98,.... etc
100x101/2 = 50x101 = 5050
1,2,...., n
1+n, 2+(n-1), 3+(n-2), .... etc.. Each of these pairs adds up to 1+n
There are n/2 such pairs....
Sum is therefore... n(n+1)/2
Here... it's even easier as you know n=100
1+100, 2+99, 3+98,.... etc
100x101/2 = 50x101 = 5050
Animis opibusque parati
I would probably avoid algebra but do it steps.
1+2+3+4+ 5+6+7+8+9 = 45
You will have 10 loads of this ie. 450 which takes care of all the units from 1 to 100.
Then you will have to cater for all the tens i.e. 10 loads of 10's (100), 10 loads of the 20's (200), 10 loads of the 30's (300). By that time a child would hopefully see the pattern and realise they will have 450 x 10 which is 4500.
Finally you have the last 100.
So 450+4500+100 = 5050.
I don't think most children, even clever ones, would come up with an algebraic formula but are much more likely to evolve a method, as above, while attempting to add them all.
1+2+3+4+ 5+6+7+8+9 = 45
You will have 10 loads of this ie. 450 which takes care of all the units from 1 to 100.
Then you will have to cater for all the tens i.e. 10 loads of 10's (100), 10 loads of the 20's (200), 10 loads of the 30's (300). By that time a child would hopefully see the pattern and realise they will have 450 x 10 which is 4500.
Finally you have the last 100.
So 450+4500+100 = 5050.
I don't think most children, even clever ones, would come up with an algebraic formula but are much more likely to evolve a method, as above, while attempting to add them all.
Small variation on herman and SVE's methods, avoiding the trickness in the middle: once you start adding
1+100 = 101
2+99 = 101
3+98 = 101
...
it's easy to keep going:
1+100 = 101
2+99 = 101
3+98 = 101
...
98+3 = 101
99+2 =101
100+1 = 101
That tells you that twice the sum of 1 to 100 is 100*101 = 10100, so the sum is half that (5050).
1+100 = 101
2+99 = 101
3+98 = 101
...
it's easy to keep going:
1+100 = 101
2+99 = 101
3+98 = 101
...
98+3 = 101
99+2 =101
100+1 = 101
That tells you that twice the sum of 1 to 100 is 100*101 = 10100, so the sum is half that (5050).
Funny!
I would just do the average x the number (but prototype it on 1..10). So 50.5 * 100 = 5050
Of course the trick is to realise that the average is 50.5 (not the more obvious answer of 50). This is why the prototype test is useful (average of 1..10 is 5.5)
How strange, the ways in which different people work it out.
I would just do the average x the number (but prototype it on 1..10). So 50.5 * 100 = 5050
Of course the trick is to realise that the average is 50.5 (not the more obvious answer of 50). This is why the prototype test is useful (average of 1..10 is 5.5)
How strange, the ways in which different people work it out.
Re: A strategy for the following Maths Question
funny enough, that was the question my DS had in his Hampton school 11 plus some years ago, and now I have taught my DD the average method mentioned above.
A similar type of question appeared in exam recently,
if 5! = 5x 4 x 3 x 2 x 1
what is
a) what is 7!
b) 10! / 9!
c) 49! / 47!
A similar type of question appeared in exam recently,
if 5! = 5x 4 x 3 x 2 x 1
what is
a) what is 7!
b) 10! / 9!
c) 49! / 47!
Re: A strategy for the following Maths Question
or say:
total = 1 + 2 + 3 + 4 + .................................+ 99 + 100 reversing gives
total = 100 + 99 + ............................................ 2 + 1 then adding
double the total = 101 + 101 + ....................................+ 101 and there's 100 of those so my total is 50 x 101
Look up Gauss - his teacher set him this to 'keep him quiet' and this is the method he came up with in a few moments
There - even some maths history to boot
total = 1 + 2 + 3 + 4 + .................................+ 99 + 100 reversing gives
total = 100 + 99 + ............................................ 2 + 1 then adding
double the total = 101 + 101 + ....................................+ 101 and there's 100 of those so my total is 50 x 101
Look up Gauss - his teacher set him this to 'keep him quiet' and this is the method he came up with in a few moments
There - even some maths history to boot