# Thread: Senior Maths Challenge question

1. ## Senior Maths Challenge question

Hello lads. So I did my senior maths challenge today, and there was a question which struck me as being difficult. I even asked my maths teacher and he couldn't solve it as fast as he could. So who can run me through how you would solve the following:

abcd + abc + bcd + acd + abd + ab + bc + cd + ad + ac + bd+ a + b + c + d = 2009

Find a+b+c+d

Anyone got any idea how to do this?

2. Originally Posted by 123ohrid
Hello lads. So I did my senior maths challenge today, and there was a question which struck me as being the sort of WTF questions. I even asked my maths teacher and he couldn't solve it as fast as he could. So who can run me through how you would solve the following:

abcd + abc + bcd + acd + abd + ab + bc + cd + ad + ac + bd+ a + b + c + d = 2009

Find a+b+c+d

Anyone got any idea how to do this?
Others may prove me wrong but I don't think there's a unique answer to this question. You'll notice that adding 1 to both sides gives

$\displaystyle (a+1)(b+1)(c+1)(d+1) = 2010.$

Some solutions

$\displaystyle a = 1, \;b = 1004, \; c = 0, \;d = 0,$

$\displaystyle a = 2, \;b = 669, \; c = 0, \;d = 0,$

$\displaystyle a = 4, \;b = 401, \; c = 0, \;d = 0,$

$\displaystyle a = 5, \;b = 334, \; c = 0, \;d = 0,$

$\displaystyle a = 9, \;b = 200, \; c = 0, \;d = 0.$

We see that $\displaystyle a+b+c+d$ changes in all these cases.

3. I bet the question said none can equal 0. If so, 2010 breaks down into 4 prime factors which furthers this conclusion - 2,3,5,67.

4. Originally Posted by Jameson
I bet the question said none can equal 0. If so, 2010 breaks down into 4 prime factors which furthers this conclusion - 2,3,5,67.
Nice observation!

5. Yeah it sayd that a b c and d are positive integers.

Also this might help, the answer is one of the following:

A)73 B) 75 C) 77 D) 79 E) 81..

So which one would it be?

I'm guessing 77 because it's the sum of all your primes? But can you try to explain why?

6. $\displaystyle (a+1)(b+1)(c+1)(d+1) = 2010$

2010 has exactly 4 prime factors -- 2, 3, 5, 67 (this means that $\displaystyle 2010 = 2\cdot 3 \cdot 5 \cdot 67$. Convince yourself that you cannot write 2010 as another product of 4 different factors.), so $\displaystyle (a+1) + (b+1) + (c+1) + (d+1) = 2 + 3 + 5 + 67 \Rightarrow a + b + c + d = 73$