You are currently browsing the tag archive for the ‘ib maths’ tag.

I’ve just made a big update to both the teacher and student resources sections:

Student resources

Screen Shot 2022-06-12 at 4.17.08 PM

These now have some great free resources for students to help them with the IB maths course – including full course notes, formula books, Paper 3s, an Exploration guides and a great mind-map.  Make sure to check these all out to get some excellent support for the IB maths course.

Teacher resources

Screen Shot 2022-06-12 at 4.18.45 PM

These now have over 25 worksheets, investigations, paper 3s, treasure hunts and more resources – both with question pdfs and markscheme pdfs.  I’ve added a lot of enriching activities that would support explorations and paper 3 style problems and also put a selection of some excellent other resources from IB teachers too.

So be sure to check these both out!


Can you find a sequence of consecutive integers that add up to 1000?

This puzzle is based on the excellent book A First Step to Mathematical Olympiad Problems – which is full of problems that could be extended to become exploration ideas.

Step 1 – arithmetic formula

Our first step is to write out what we want:

a + (a+1) + (a+2) + … (a +n) = 1000

next we notice that the LHS is an arithmetic series with first term a, last term a+n and n+1 terms.  Therefore we can use the sum of an arithmetic sequence formula:

Sn = 0.5n(u1 + un)

Sn = 0.5(n+1)(a + a+n) = 1000

Sn = (n+1)(2a+n) = 2000

Step 2 – logic

However, we currently have 2 unknowns, n and a, and only 1 equation – so we can’t solve this straight away.  However we do know that both a and n are integers – and n can be taken as positive.

The next step is to see that one of the brackets (n+1)(2a+n) must be odd and the other even (if  n is odd then 2a + n is odd.  If n is even then n+1 is odd).   Therefore we can look at the odd factors of 2000:

Step 3 – prime factorisation

Using prime factorisation: 2000 = 24 x 5³

Therefore any odd factors must solely come from the prime factor combinations of 5 – i.e 5, 25 and 125.

Step 4 – trial and error

So we now know that either (n+1) or (2a+n) must be 5, 25, 125.  And therefore the other bracket must be 400, 80 or 16 (as 5 x 400 = 2000 etc).  Next we can equate the (n+1) bracket to one of these 6 values, find the value of n and hence find a.  For example:

If one bracket is 5 then the other bracket is 400.

So if (n+1) = 5 and (2a+n) = 400 then n = 4 and a = 198.

This means that the sequence: 198+199+200+201+202 = 1000.

If (n+1) = 400 and (2a+n) = 5 then n = 399 and a = -197.

This means the sequence: -197 + -196+ -195 … + 201 + 202 = 1000.

We follow this same method for brackets 25, 80 and 125,16.  This gives the following other sequences:

28+29+30+…+51+52 = 1000

-54+-53+-52+…+69+70 = 1000

-27+-26+-25+…+51+52 = 1000

55+56+57+…+69+70 = 1000

So with a mixture of mathematical formulae, prime factorisation, logic and trial and error we have our solutions.  A good example of how mathematics is often solved in reality!

Website Stats



All content on this site has been written by Andrew Chambers (MSc. Mathematics, IB Mathematics Examiner).

New website for International teachers

I’ve just launched a brand new maths site for international schools – over 2000 pdf pages of resources to support IB teachers.  If you are an IB teacher this could save you 200+ hours of preparation time.

Explore here!

Free HL Paper 3 Questions

P3 investigation questions and fully typed mark scheme.  Packs for both Applications students and Analysis students.

Available to download here

IB Maths Super Exploration Guide

A Super Exploration Guide with 168 pages of essential advice from a current IB examiner to ensure you get great marks on your coursework.

Available to download here.

Recent Posts

Follow IB Maths Resources from Intermathematics on