# Thread: World's Most Difficult Math Problems

1. ## World's Most Difficult Math Problems

<font color=blue> <font face="Comic Sans MS">Maybe, maybe not, but here's a couple that should keep you busy for a while...</font face=comic> </font color=blue>

A starter question...
There is a five-digit number A. With a 1 after it, it is three times as large as with a 1 before it. What is the number?

Now, the tournament question...
Find a square which remains a square if it is decreased by 5 or increased by 5.

2. ## Re: World's Most Difficult Math Problems

Starter:
<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>Placing 1 after A yields 10*A+1.
Placing 1 before A yields 100000+A (since A is a 5 digit number)
10*A+1 = 3*(100000+A)
7*A = 299999
A = 42857</font color=yellow></span hi>

3. ## Re: World's Most Difficult Math Problems

Tournament: seems to be a Tricky puzzle!
<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>If you mean the square of a whole number, there is no solution. The only squares of whole numbers that differ by 5 are 2^2 =4 and 3^2 = 9. If you remove the restriction that it must be squares of whole numbers, there is an infinite number of solutions, since every positive number is the square of something (namely, the square root of that number). If you also allow complex numbers, even negative numbers are squares. In that case 9 = 3^2, 4 = 2^2, -1 = i^2 where i is the imaginary number representing the square root of -1.</font color=yellow></span hi>

4. ## Re: World's Most Difficult Math Problems

Okay, I'll give you the correct solution is not an integer, but it is also not an imaginary number. I was unable to ask the question and make it clear the type answer I was seeking, so consider the following example:

If I had asked, "Give me a square that remains a square when increased or decreased by the number 6", a correct response would have been -

<font color=blue>The number is 6 1/4, being the square of 2 1/2. 1/4 and 12 1/4 are the squares of 1/2 and 3 1/2.</font color=blue>

So, in that fashion, there is a good solution for the original question...

5. ## Re: World's Most Difficult Math Problems

Hans - Nice work on both puzzles. I, too, was able to solve the first one with relative ease. But that second one had me going for over two hours, even with Excel at my perusal. In the end, I did not solve it. I figured trial & error to be the only solution as algebra wasn't working... If time permits, maybe you could share the steps involved in solving it.

Have a great day <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16>

6. ## Re: World's Most Difficult Math Problems

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>We are looking for three fractions; by multiplying the numerator and denominator of each by a suitable number, way may assume that the denominator of all three is the same number.

Say the number + 5 is (a/n)^2, the number itself is (b/n)^2 and the number - 5 is (c/n)^2, where a, b, c and n are positive integers.

From (a/n)^2 - (b/n)^2 = 5 and (b/n)^2 - (c/n)^2 = 5 we can deduce a^2 + c^2 = 2*b^2 and a^2 - c^2 = 10*n^2, or (a^2 + b^2)/2 is the square of a whole number and (a^2 - c^2)/10 is the square of a whole number.

I created a spreadsheet with the numbers a and c (where a > c), and tested for which a and c the above is true. the only numbers under 50 for which this is the case are a = 49 and c = 31, from which we can compute b = 41 and n = 12 using the equations above.

See attached zipped spreadsheet.</font color=yellow></span hi>

7. ## Re: World's Most Difficult Math Problems

It would be nice if the puzzle stated what kind of numbers is allowed. In most math puzzles, if not specified explicitly, it is assumed that solutions must be whole numbers.
<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>The square of 41/12</font color=yellow></span hi>

8. ## Re: World's Most Difficult Math Problems

Way Cool ! <img src=/S/clapping.gif border=0 alt=clapping width=19 height=23>

I got as far as:
<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>x^2 + 5 = u^2 and x^2 - 5 = v^2</font color=yellow></span hi>
Then:
<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>u^2 - v^2 = 10</font color=yellow></span hi>
But that's where I hit the proverbial brick wall and thought that the trial & error approach would be the only place to go... <img src=/S/brickwall.gif border=0 alt=brickwall width=25 height=15>
Thanks for showing me how.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•