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Calculator solving some mathematical competition task

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Yesterday I was asked to help with some math tasks. They could be solved head on with direct calculation or with enumerative technique, but that will take too long for a person. As I was too lazy to find a smart way, I gave this task to the computer by creating some calculators. It does that definitely faster.

So, the first task is going like this: Consider the alternating sum 1 3 – 5 7 + 9 11 – 13 15 + … – 2005 2007 + 2009 2011. What is their sum?
The calculator will find the sum simply following the procedure below. :)

### Odd numbers calculation

Sum

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But in fact, this problem is solved quite easily, and the answer matches the one counted by the computer. In short, it all can be replaced with
$1 * 3 - 5 * 7 + 9 * 11 - 13 * 15 + ... - 2005 * 2007 + 2009 * 2011$$(2^2 - 1) - (6^2 - 1) + (10^2 - 1) - (14^2 - 1) + ... - (2008^2 - 1) + (2010^2-1)$
All 1 are reduced, except the last one
$2^2 - 6^2 + 10^2 - 14^2 + ... - 2008^2 + (2010^2-1)$

calculating a couple of terms, we can see that the difference between two consecutive terms equal to -32n, where n = 1,3,5 etc. until 501
$32$
(-1) + 32(-3) + ... + 32(-501) + (2010^2-1)
if we put 32 out of the brackets, then inside there will be an arithmetic progression, the sum of which can be calculated quickly using Arithmetic progression :). Well, and then multiply by 32 and subtract from the remaining term of 2009
2011

The second task: We have a grandfather who is older than 80 (but younger than 150)/ Today he can tell his grandchildren who have different ages: «The product of our three ages is the sum of squares of our ages». Determine the age of the grandfather. If you simplify the wording - it is necessary to find three numbers whose product is equal to the sum of their squares. Calculator that solves this problem with enumerative technique is below :)

### Products and sums

1st number

2nd number

3rd number

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To be honest, I couldn't think of the way to solve this without a calculator. Probably you just need to reduce the number of options by, for example using the fact that the product of two grandchildren ages should be slightly more than the age of the grandfather. Perhaps there is some more elegant way to solve this - share if you know for sure.

The third task: 2011 + BON + JEU = MATH
In this rebus, each letter stands for a digit from 0 to 9. Two different letters are always replace two different digits and no number will start from 0. Determine the maximum value of MATH. Calculator that solves this problem with enumerative technique is below :) By the way it takes a lot of time (for a computer, of course)

### Nonrecurrent numbers

First term

Second term

Sum

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In principle, this problem can also be solved mentally. It is clear that B, J, M and A are located immediately. With the rest, you just need to fiddle a bit.
That's cheating of some sort