# The "I am very bored" thread!

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Ok,
I will admit it. I am really bored! I am desperately looking for ways to waste my time, but couldn't find any, so I decided to start another pointless thread!

I hope that this thread will become some sort of a "boredom resource" with lots of small ideas for things to do if you are bored! Any suggestions??

Aangemeld of registreer om reacties te plaatsen

Ok, I will start...

1. Find two numbers in the form 1/x such that their sum is 19/94. (i.e. find x,y such that 1/x + 1/y = 19/94)

2. Solve the equation
ABCDE
*
4
______
EDCBA
where each letter represents a different digit!

1. 1/x + 1/y = (x+y)/(x*y)
94 = 2*47. 2 and 47 are prime, 2+47 != 19, so there's no solution, I'd say...

2.

```[~\$] more abcde.c
#include <stdio.h>
int main()
{
int a,b,c,d,e;
for(a=0; a<10; a++)
for(b=0; b<10; b++)
for(c=0; c<10; c++)
for(d=0; d<10; d++)
for(e=0; e<10; e++)
if(a != b && a != c && a != d && a != e &&
b != c && b != d && b != e &&
c != d && c != e &&
d != e &&
(a + 10 * b + 100 * c + 1000 * d + 10000 * e) ==
((e + 10 * d + 100 * c + 1000 * b + 10000 * a) * 4))
{
printf("a=%d b=%d c=%d d=%d e=%d\n",a,b,c,d,e);
printf("%d * 4 = %d\n",
(e + 10 * d + 100 * c + 1000 * b + 10000 * a),
(a + 10 * b + 100 * c + 1000 * d + 10000 * e));
}
}
[~\$] gcc -o abcde abcde.c
[~\$] ./abcde
a=2 b=1 c=9 d=7 e=8
21978 * 4 = 87912
[~\$] _```

Cheat:
x = 9.5 + i * Sqrt(15)/2
y = 9.5 - i * Sqrt(15)/2

For solutions to boredom, go here. ;)

Yay! I could deceive two smart guys! Actually the 19/94 thing HAS a solution No cheats or anything!

Oh, and good job with the ABCDE thing Nice piece of code

Yay! I could deceive two smart guys!
Hey, you caught me on sunday (well...) morning before I had my coffee. Go figure
Actually the 19/94 thing HAS a solution No cheats or anything!
Prove it
Oh, and good job with the ABCDE thing Nice piece of code
Thanx! I call it 'brute force'

Actually, the solution is: y = 94*x / (19*x - 94)
With the smallest integer solution for x=5 we have 1/5 + 1/470 = 19/94

Actually, the solution is: y = 94*x / (19*x - 94)
With the smallest integer solution for x=5 we have 1/5 + 1/470 = 19/94

Correct! Good Job! I am impressed

Now does any one have any others of the same kind?

Ok, here is another one...

A certain whole number whose last digit is 7, has the property that in order to multiply it by 7, all that is needed it to take the 7 from the right end and place it at the beginning. What is this number?

This one is specifically for AuroraMSX, (because brute force will never work here )

from the RIGHT end to the beginning?

Where is the right end, and where is the beginning?

lol

The right end is... well... at the far right
The beginning is at the far left

Ok the solution is quite easy
We HAVE to multiply by 7
AND the number ends with a 7
Now take a calculator and multiply ANY number ending with 7 by 7.
Whats the number before the last one? 9
Why

7*7=49

Ok, so far the easy part. Now I hope I can explain this clearly.
Now we know the number we are looking for ends with 97
Now we separate the 7, because we already did our math on that number

This leaves us 90
now again we have to multiply by 7
so 90*7=630

49 + 630 = 679

Now follow this: the number ends with 797
Why?
the last 7 is the 7 WE KNEW the number HAD to end at
the 9 is the number we got from multiplying 7*7
the 7 is the number we got from multiplying 90*7

we can keep this pattern going, this would be the result:

7*7=49 ;this is our first 7 (first number from the right)

90*7=630 ;this gives us our 9 (second number from the right)
630+49=679

700*7=4900 ;this gives us our 7 (third number from the right)
4900+679=5579

5000*7=35000 ;this gives us our 5 (fourth number from the right)
35000+5579=40579

00000*7=0 ;and this gives us our 0 (fifth number from the right)

This means the number = 05797

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