11 Plus Exams Forum

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.

suspect you were on the right track because if you keep doing that:...

there will be
99+1
98+2
...

36+64
...

right up to

49+ 51

so that means there are 49 of these plus the 50 left on its own and the hundred...

4900+100+50 = 5050.

OK mid post surgery brain addled - probably got it wrong,,

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

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.

I prefer Herman's method.

Reminds me of number bonds, that kids are taught.

steve

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).

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.

There are 50 pairs of 101 using the 1 + 100, 2 + 99 method. So 101 times 50.

I use 1-10 as an assessment tool to see which methods children come up with.

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!

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