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Chocolate 2010

Age 14 to 16
Challenge Level Yellow star
  • Problem
  • Getting Started
  • Student Solutions
  • Teachers' Resources

There were a significant number of correct solutions to this problem and one of the most well expressed is given below. A number of younger members sent examples of the "trick" working and it was good to see you investigating what was happening. The solution chosen tries to explain why. Some of you needed to dig a little deeper! Well done to all of you who successfully solved this.

David of Colyton Grammar School gave the most comprehensive and well explained solution. Two other good solutions were received from Charlene of Sekolah Menengah Pengiran Jaya Negara Pengiran Haji Abu Bakar and Vu of School B6.

We take the number of pieces of chocolate that we can have as x.

First, we multiply it by 2, so we have 2x.

Adding 5 to it, we now have 2x + 5

Multiplying it by 50 we have 50*(2x + 5) = 100x + 250

Finally adding 1753 or 1754 to the equation on the previous line, we get 100x + 250 + 1753 or 100x + 250 + 1754, which, after subtracting year of birth (take it as y) is either

1. 100x + 2003 - y
OR
2. 100x + 2004 - y

Multiples of 100 always end in double-zero digits, so what the tens and units digits are depends on the value of 2003 - y or 2004 - y (this depends on the year you last celebrated your birthday), which gives your age! Note that the trick does not work for centenarians as their age would exceed two digits.

We also see that the hundreds digit is equal to x, provided that x is not larger than 10.


To start with you are asked to choose an integer between 1-10 which I will call "x".

WE HAVE: x
The first calculation is to multiply by 2.
SO WE HAVE: 2x
Then we are told to add 5.
SO WE HAVE: 2x + 5
Following that is an instruction to multiply by 50.
GIVES US: 100x + 250
Then depending on whether or not you have had your birthday yet this year, you are asked to add 1753 or 1754.
GIVING: 100x + 2003 OR 100x + 2004
The last stage is to subtract the 4 digit year you were born.
GIVING: 100x + 2003/4 - 19yy (yy is the year you were born)

This gives us the answer we were promised of "xzz" (zz is your age)

Now it is clear to see how the problem works. As soon as you have multiplied by 2 and 50 you will have 100x. For the remainder of the instructions this remains and will be 100, 200, 300... 900.

The last two parts are for getting the remaining two digits of the answer, your age. It is simply done by subtracting the year you were born from the year in which you last had a birthday so either from 2003 or 2004. When you added 5 in step 2 and then multipled by 50 it meant you already had 250 so to get up to 2003/4 you had to add 250 less hence the addition of 1753 or 1754.

Potential Problem:

The obvious problem is that this will only work during 2004. So in previous or future years the instructions need to be adapted. In place of 1753 you need the current year minus 251 and in place of 1754 you need the current year minus 250.

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The nth term of a sequence is given by the formula n^3 + 11n. Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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