Mathematical Side Show

Soroban · #1 $Old$ March 22nd, 2009, 06:23 AM

Mathematical Side Show

Here are some examples of unusual mathematical trivia.
Do you have any to contribute?

$\begin{array}{ccc}1,\!001 &=& 7\cdot11\cdot13 \\ 10,\!001 &=& 73\cdot137 \\ 1,\!000,001 &=& 101\cdot 9901 \\ 10,\!000,\!001 &=& 11\cdot909091 \end{array}$

$\begin{array}{ccc}111 &=& 3\cdot37 \\ 1,\!111 &=& 11\cdot101 \\ 11,\!111 &=& 41\cdot271 \\ 111,\!111 &=& 3\cdot7\cdot11\cdot13\cdot37 \\ 1,\!111,\!111 &=& 239\cdot 4649 \\ 11,\!111,\!111 &=& 11\cdot73\cdot101\cdot137 \\ 111,\!111,\!111 &=& 9\cdot37\cdot333667 \end{array}$

$\begin{array}{ccc}\dfrac{37^3+13^3}{37^3+24^3} \;=\;\dfrac{37+13}{37+24} \\ \\[-2mm] \dfrac{3^4 + 25^4+38^4}{7^4+20^4+39^4} \;=\;\dfrac{3+25+38}{7+20+39} \end{array}$

$\begin{array}{ccc}168 \times 861 &=& 294 \times 492 \\ 276 \times 672 &=& 384 \times 483 \\ 1476 \times 6741 &=&2583 \times 3852\\ 2556 \times 6552 &=& 3744 \times 4473 \end{array}$

A classic "oldie"

$\text{Solve: }\:a^b \:=\:b^a\,\text{ in rational numbers.}$

$\text{Most people know that: }\:2^4 \:=\:4^2$
$\text{But very few of us know that: }\:\left(\frac{27}{8}\right)^{\frac{9}{4}} \;=\;\left(\frac{9}{4}\right)^{\frac{27}{8}}$

Soroban · #2 $Old$ March 23rd, 2009, 07:57 AM

Squares containing all ten digits

$32,\!043^2 \:=\:1,\!026,\!753,\!849$

$32,\!286^2 \:=\:1,\!042,\!385,\!796$

$33,\!144^2 \:=\:1,\!098,\!524,\!736$

$35,\!172^2 \:=\:1,\!237,\!069,\!584$

$39,\!147^2 \:=\:1,\!532,\!487,\!609$

$45,\!624^2 \:=\:2,\!081,\!549,\!376$

$55,\!446^2 \:=\:3,\!074,\!258,\!916$

$68,\!763^2\:=\:4,\!728,\!350,\!169$

$83,\!919^2 \:=\:7,\!042,\!398,\!561$

$99,\!066^2 \:=\:9,\!814,\!072,356$

Unusual Summation

Most of us are familiar with this identity:

$(1+2+3+ \hdots + n)^2 \:=\:1^3+2^3+3^3+\hdots+n^3$

But how about this one?

$2(1+2+3+\hdots +n)^4 \:=\:(1^5+2^5+3^5+\hdots +n^5) + (1^7 + 2^7 + 3^7 + \hdots + n^7)$

The fourth power of the sum is the average of
the sum of the fifth powers and the sum of the seventh powers.

Sums of Consecutive Squares

$3^2+4^2 \:=\:5^2$

$10^2+11^2+12^2 \:=\: 13^2 + 14^2$

$21^2 + 22^2 + 23^2 + 24^2 \:=\:25^2 + 26^2 + 27^2$

$36^3 + 37^2 + 38^2 + 39^2 + 40^2 \:=\:41^2 + 42^2 + 43^2 + 44^2$

$55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 \:=\:61^2 + 62^2 + 63^2 + 64^2 + 65^2$

Yes, there is a pattern here.
Can you find the general equation?

Chop Suey · #3 $Old$ March 23rd, 2009, 09:40 AM

What I got isn't so pretty:
$\underbrace{(2a^2-3a+1)^2 + (2a^2 - 3a + 2)^2 + \ldots + (2a^2 -2a)^2}_{a~\text{terms}}$
$= \underbrace{(2a^2-2a+1)^2 + (2a^2-2a+2)^2 + \ldots + (2a^2 - a -1)^2}_{a - 1~\text{terms}}$

Is there another way?

Soroban · #4 $Old$ March 24th, 2009, 05:00 AM

Hello, Chop Suey!

Mine is also unattractive . . .

$\underbrace{[n(2n+1)]^2 + [n(2n+1) + 1]^2 + [n(2n+1) + 2]^2 + \hdots + [n(2n+1) + n]^2}_{n+1\text{ terms}}$

. . $= \;\underbrace{[n(2n+1) + (n+1)]^2 + [n(2n+1) + (n+2)]^2 + \hdots + [n(2n+1) + 2n]^2}_{n\text{ terms}}$

Left side:
The first term is: $n(2n+1)$ squared.
. . Add the next $n$ squares.

Right side:
Add the next $n$ squares.

This is not much of an improvement: . $\sum^n_{k=0}\bigg[n(2n+1) + k\bigg]^2 \;=\;\sum^{2n}_{k=n+1}\bigg[n(2n+1)+k\bigg]^2$

Thank you, mods, for restoring my post.
I thought I could improve it, but I really botched it.
.