Maths trivia

[SydneyGeek]

Just got reminded of something today...

1^3 + 2^3 + 3^3 = (1 + 2 + 3)^2
Keeps working as you extend the series.

Anyone else got a tidbit?

Denis

 

[Joe4]

One of my favorites is that is you add up the digits of any number, and that sum is divisible by 9, then that number is divisible by 9, i.e.

1485
1 + 4 + 8 + 5 = 18
18 / 9 = 2

1485 / 9 = 165

 

[sulakvea]

Take any integer, sum up ODD digits, and then EVEN digits. subtract one from another. if the result is divisible by 11, then the original number is divisible by 11 as well. say - 121. sum of odds =1+1=2. sum of evens =2. 2-2=0, thus 121 is divisible by 11.

313010487215450416255248 - sum of odds =35, sum of evens =46. divisible by 11

 

[Fazza]

A simple pattern that I like, especially I think because even young children can 'see it'.

1 * 1 = 1
11 * 11 = 121
111 * 111 = 12321
1111 * 1111 = 1234321
11111 * 11111 = 123454321
111111 * 111111 = 12345654321

It is great to then ask about 111111111 * 111111111
Even youngsters can respond with 12345678987654321

 

[MorganO]

As a youth learning math I liked this concept with the 9's:

1 * 9 = 09 | 0 + 9 = 9
2 * 9 = 18 | 1 + 8 = 9
3 * 9 = 27 | 2 + 7 = 9
.
.
.
9 * 9 = 81 | 8 + 1 = 9
10 * 9 = 90 | 9 + 0 = 9
11 * 9 = 99 | 9 + 9 = 18 | 1 + 8 = 9

 

[Sandeep Warrier]

The easiest way to get the square of any number ending with 5...

25^2 => take the square of the last digit i.e. 5 and store it. 5^2 = 25.

Now take the 1st digit and multiply it with the number obtained by incrementing the digit by 1 => 2 * (2+1) => 2*3 => 6. Now concatenate this with 25 => 625 = 25^2

Similarly for 35^2 = (3*4) 25 => 1225

etc.
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[Gerald Higgins]

Joe4 - you can extend the 9 example.

If the digits sum to 3, 6 or 9, then the number is divisible by 3.
If the digits sume to 3, and the original number is even, then the number is divisble by 6.
There are probably more of these.

 

[pgc01]

One thing I remember from highschool that I find interesting is that given 2 numbers, if you know their sum S and their product P, you can find them by solving the equation:

x² - Sx + P = 0

 

[Oaktree]

The one that still makes my head hurt is that there are more integers than rational numbers.

 

[Joe4]

The one that still makes my head hurt is that there are more integers than rational numbers.

What? How can that be? I need some clarification on this one!

Isn't every integer a rational number and not every rational number is an integer?

Or is it the argument that integers are infinite, so how can there be anything more than infinity? If that is case, you might be able to make the argument that there are NOT more rational numbers than integers, but I don't think you can say that there are more integers than rational numbers.