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/sci/ - Science & Math

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3498847 No.3498847 [Reply] [Original]

Greeting /sci/, new to this board, changing my ways.
I had my first day of univeristy the other day, and a maths lecturer blew my mind. I'm wondering if any of you have the rule/equation/whatever it is.
Here's the scenario, select 4 random numbers.
Lets choose (for argument's sake) 5482.
Reverse that number. 2845.
Minus the smaller number from the larger.
5482-2845=2637
Times the answer by 3.
2637*3=7911
Circle any single number from the answer.
Lets pick 7.
Now with the numbers 9,1,1, the lecturer would "read your mind" and come up with the number 7.
He did this for 10 students, guessing them all correct.
Can anyone explain how it was done? It's been driving me crazy!

>Pic not related, all I have on laptop atm.

>> No.3498878

9-1-1=7

>> No.3498883

basic arithmetic

>> No.3498886
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3498886

>>3498847

You call this trolling?

>> No.3498890

lol your lecturer did this for 10 students? What a waste of time. What community college do you go to?

>> No.3500153

>>3498878
Do it with another, pick your own 4 digits.

>> No.3500155

>>3498890
>Implying 2 minutes is a lot of time in a 2 hour long lecture.

>> No.3500222

1828

>> No.3500227

8281-1828=6453

i chose read my mind!

>> No.3500261

OP here, you're all faggots. I got it by myself.
7834-4387=3447
3447*3=10341
10(3)41 (3 is the number you're trying to guess)
1+0+4+1=6
Skip to the next factor of 9, so from 6, we go to 9.
9-6=3 (The chosen number!)

Another one just for good measure.
6895-5986=909
909*3=2727
2(7)27
2+2+7=11
Next factor of 9 being 18.
18-11=7

>> No.3500675

>>3500261
>Now why does that work?

>> No.3500743

You are asked to chose a number on the form
<span class="math">c = a10^3 + b10^2 + c10 + d[/spoiler]
Note that the 'inverse' is on the form:
<span class="math">c^{'} = d10^3 + c10^2 + d10 + a[/spoiler]
therefore the difference:
Assume for the sake of the argument that a>d.

<span class="math">c - c^{'} = (a-d)10^3 + (b-c)10^2 + (c-b)10 + d-a[/spoiler]

Notice this number is divisible by three. (simple divisibility rule of adding together the coefficients).

Then he asks you to multiply the answer by three. This means your number is now divisible by 9!.

Notice 9 divisibility rule; number is divisible by 9 only if the sum of integers of the number is divisible by 9.
All your prof has to do is simply add all the remaining numbers together, and see which number he needs to add to make it divisible by 9.
I'll let you discover on your own how (if) he reacts when the digits he receives already are divisible by 9. Then there are two options, last digit is 0 or 9.
Go /sci/

>> No.3500758

ripoff from that mathemagician guy