Add up all of the numbers from
1 to 50

Show a hint

A Mathematics Lesson Starter Of The Day

Topics: Starter | Algebra | Number | Problem Solving | Sequences

• Kerei James, NZ
•
• It is an excellent resource.
• S.W, Redbourn Junior
•
• I would have this starter up when the children enter the room. They work out the answer on white boards and as soon as they have finished they write their initials against the highest number available in a list of numbers from 1 to 10 on another board - the TOP TEN list. When the TOP TEN is full up we stop and look at the answer together.
• Nicola, Neilston Primary School
•
• My class counted them all up and got the answer 703 then double checked it with a calculater and they were correct.
• Class A, Furness Vale Primary School
•
• We get 1176 doubled checked on a calculator.
•
• The correct answer is 820 like it says at the bottom. The formula you use is for consecutive sum.
sum=n*(n+1)/2
sum = 40(40+1)/2
sum = 1640/2
sum = 820.
• Ray Dunne, Ireland
•
• Answering the question as written ,assuming there is no word play and using universal basic addition I concur with 1176. I would prefer your method though as my sallery would be greatly increased.
• Ray Dunne, Ireland
•
• While Cody from Canada has the formula correct he has inputted the wrong information . if you change 40 and 41 for 48 and 49 then you get 1176 which is the correct answer.
• Jonathan, Wales
•
• The triangular number formula is correct.
However I would like to point out that the numbers used sometimes differ hence why different people are saying different results.
• Mrs Zaker - Teacher Of Maths, Tolworth Girls' School, London UK
•
• Using Carl Friedrich Gauss's method: we have 22 pairs that add to 46 (half of 45 is 22) PLUS the 23rd number (which is 23)
45 + 1 = 46
44 + 2 = 46
43 + 3 = 46
etc.
so 22 x 46 = 1012
add on 23 that gives 1035.
• Transum,
•
• Please note that each time this page is loaded the number of numbers in the question changes. Consequently the solutions suggested here in the comments will refer to different variations of this starter. Thanks Jonathan for pointing this out. Thanks also for the comments and explanations of the methods you used. keep them coming!
• Angus Dresner, O.K.C.M.S
•
• Just today I found this method myself. If you half the biggest number which in my case is 56 you get 28. then add .5 to get 28.5. Multiply the two numbers together to get 1596.
• Uzma, Barking
•
• Hi all
I found it 1176 . I want to tell you the way I followed. I think it is the easiest method. First add all th 10s in the sum. Then find how many 1s,2s.......8s in the sequence. These numbers coming 5 times in the sequence, so multiply each number from 1-8 with 5.and add all in the sum of 10s. Finally add 9x4, as 9 appear only 4 time in the sequence.
• G Morkel, Albany Junior High School NZ
•
58(58+1)/2
OR, you split the numbers in half and pair them up as 58+1, 57+2, 56+3, etc, you will then have 29 groups of 59 which is 1711.
• Alan, Australia
•
The question did NOT say to just add the whole numbers
if just whole numbers, the answer is 990 ( 45 x 22 ).
• James Streeter, Luton
•
• None of you are correct!
It is 1081
There are 23 pairs of 47
46+1
45+2
44+3
43+4
etc...
Until you reach the middle - which is
24+23
Exactly twenty three pairs!
23 x 47 = 1081
Definitely correct.
• Eamon,
•
• It has already been pointed out that the 2nd number changes via a random number generator when the page is loaded/reloaded. From what I can see, the random number generator has 3 rules
- The number has to follow this expression: ൦ is less than or equal to n" and "n is less than 60"
- The number has to be an integer
- The number has to be even
With these rules, if you decide to pair numbers together as your method of choice, there should be nothing remaining as a "stray" or "middle", unless you are going to attack the wording choice of "number" instead of "integer".
• Rey, Vietnam
•
• The sum of each pair of numbers is 53. There are 26 pairs. 26 x 53 = 1378. So, the sum is 1378.
• Ramu Chikkala, Visakhpatnam
•
• Suppose if we want to find the sum of first 54 terms we can formula n(n+1)/2 or we can the forumula n/2(a+b) where 'a' and 'b' are first and last term of the sequence
so sum of first 54 natural numbers = 54/2 (1+54)
= 27 x 55
= 1485.

How did you use this starter? Can you suggest how teachers could present or develop this resource? Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for Maths teachers anywhere in the world.

Previous Day | This starter is for 26 February | Next Day

Note to teacher: Doing this activity once with a class helps students develop strategies. It is only when they do this activity a second time that they will have the opportunity to practise those strategies. That is when the learning is consolidated. Click the button above to regenerate another version of this starter from random numbers.

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Maths T-Shirts

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