Odd Differences

The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.

Factorial

How many zeros are there at the end of the number which is the product of first hundred positive integers?

Rachel's Problem

Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!

Fracmax

Stage: 4 Challenge Level:

There were a number of good solutions to this problem and I have picked several to illustrate the different approaches adopted.

First Fred and Matt from Albion Heights sent in a solution, which I rather liked. They have not explained why they did not try 1/3 as the first fraction or give full reasoning for why they knew they had a largest value, but the argument as far as it goes is a good one. I have added some examples below to illustrate what they said in a little more detail. I have included a solution based on the one from Tom or STRS, which was similar to that of Curt from Reigate School . Andrei, of Tudor Vianu College, also looked for solutions using a spreadsheet and sent in a program in C++ to search for soultions. I have included this below.

Well done to all of you.

First Fred and Matt's approach to get us started::

To crack the puzzle we tried using algebra but it did not work so we took this problem from a different approach.
Our answer to this puzzle is p=2 q=3 and r=7 and this would end up like this 1/2+1/3+1/7=41/42 .
We also tried a number of other combination but none of them was as close as this one and this is how we were thinking:
If one of the fractions is 1/2 then this puzzle would be slightly easier. Then we had to try the other two numbers. Of couse the total of the other two should be less than 1 otherwise this puzzle would equal or be more than one so the second number has to be 1/3 1/4 1/5 .... and so on.
If you take1/3 for one of the fractions, the last number has to be 1/7 for the closest result (the bigger the denominator is the smaller the number is).
If you take 1/4 then the last fraction would be 1/3 for the closest result.
Then we figure out when we use 1/2 then all we have to do is find two other numbers that add up to less than 1/2 and we know that 1/3 and 1/6 add to 1/2 so instead of 6 we used 7 and this made the answer smaller because the denonminator is bigger.

To make a fraction as large as possible you need to make the denominator as small as possible. The smallest a denominator can be is $2$. All the denominators cannot be $2$ because:

$$\frac{1}{2} + \frac{1}{2} + \frac{1}{2}= \frac{3}{2} \mbox{ which is }> 1$$

The next we can try then is

$$\frac{1}{2} + \frac{1}{3} + \frac{1}{2}= \frac{4}{3} \mbox{ which is }> 1$$

Then try the next largest value, which is:

$$\frac{1}{2} + \frac{1}{3} + \frac{1}{3}= \frac{7}{6} \mbox{ which is }> 1$$

This is getting closer to $1$ but is still greater than $1$

Try denominators of $2$, $3$ and $4$ giving:

$$\frac{1}{2} + \frac{1}{3} + \frac{1}{4}= \frac{13}{12} \mbox{ which is }> 1$$

Continuing in this way, the first set of fractions that give a total less than $1$ and with the smallest possible denominators (and largest sum) is:

$$\frac{1}{2} + \frac{1}{3} + \frac{1}{7}= \frac{41}{42} \mbox{ which is }< 1$$
I will use the notation $(p,q,r)$ when referring to set of values for the variables.

First I will eliminate the '3rd dimension' to this problem by proving that p = 2. To simplify the explanation I will always have p as the lowest variable, this is OK as p q and r are interchangeable without affecting the sum.

If $p = 3, q > = 3$ and $r > =3$ as $p$ is the lowest variable. The greatest sum in this case is: $(3,3,4)= 11/12$ as this is the minimum value each variable can take, therefore maximising the reciprocals. $(3,3,3)$ gives $1$ exactly which is not permissible as the sum must be $< 1$.

For p > = 4, the greatest sum will be given by $p, q$ and $r$ all being equal ie $(p,p,p)$, as this minimises them (they cannot be higher than $p$ as that is the lowest). The sum will be:

$$3 \times( 1 / p ) = 3 / p$$

As $p > = 4$, this cannot exceed $3/4$. Now consider the case $(2,3,7)$. The sum here is $41/42$, which is greater than the maximum established for $p = 3$ and $p \geq 4$. As $p$ is a positive integer, and $p = 0, p = 1$ both invalidate the condition that the sum must be lower than $1$, it is proven that $p = 2$.

The solution is $(2,3,7)$, or if there is a one greater still it too must occur where $p = 2$.

Now I will prove that $(2,3,7)$ in fact gives the maximum.

It is proven that $p = 2$, therefore:

\begin{eqnarray} 1/2 + 1/q + 1/r &< & 1\\ 1/q + 1/r &< & 1/2\\ q + r &< & qr/2\\ 2q + 2r - qr &< & 0\\ qr - 2q - 2r&> & 0\\ q(r - 2) - 2r &> & 0\\ q(r - 2) &> & 2r\\ q &> & (2r)/(r-2) \end{eqnarray}

If there is a solution better than that of (2,3,7), the following inequality must hold true:

\begin{eqnarray} 1/2 + 1/q + 1/r &> &41/42\\ 1/q + 1/r&> & 10/21\\ 21q + 21r &> & 10qr\\ 21q + 21r - 10qr &> & 0\\ 10qr - 21q - 21r &< & 0\\ q(10r - 21) - 21r &< & 0\\ q(10r - 21) &< & 21r\\ q &< & (21r)/(10r-21)\\ \end{eqnarray}

So it is established that

$$(2r)/(r-2) < q < (21r)/(10r-21)$$

Now I will show there are no solutions that will match this inequality. Here are some individual cases:

r = 3: 6 < q < 7

r = 4: 4 < q < 4.421...

r = 5: 3.333... < q < 3.621...

r = 6: 3 < q < 3.231...

r = 7: 2.8 < q < 3

r = 8: 2.666... < q < 2.739

None of these inequalities has an integer solution. As $r$ tends to infinity $(2r)/(r-2)$ tends to $2$ and $(21r)/(10r-21)$ tends to $2.1$. Therefore the upper bound will not exceed $3$ for $r > 8$, so the inequality will continue to fail as $q \geq 3$.

Therefore the inequality $(2r)/(r-2) < q < (21r)/(10r-21)$ has no integer solutions for $q \geq 3, r \geq 3$.

Therefore as shown earlier there is no case for $p = 2$ that yields a greater sum than the case $(2,3,7)$.

So

$p=2$
$q=3$
$r=7$

Andrei's program (I haven't tested it!):

//fracmax
\#include< iostream.h>
long double p,q,r,p1,q1,r1,max1;

void main() \{

max1=0;
for(p=1;p< =20;p++) \{
\hspace{10pt} for(q=1;q< =20;q++) \{
\hspace{20pt} for(r=1;r< =20;r++) \{
\hspace{30pt} if((1/p)+(1/q)+(1/r)< 1 \&\& 1/p+1/q+1/r> max1) \{
\hspace{30pt} max1=(1/p)+(1/q)+(1/r);p1=p;q1=q;r1=r;
\hspace{30pt} \}
\hspace{20pt} \}
\hspace{10pt} \}
\}
cout< < "p= "< < p1< < "; q= "< < q1< < "; r= "< < r1< < "; maximum= "< < max1;
\}