Modular arithmetic

patchy1 · 2016-05-17 01:00:09

This is the information I have... But I am struggling to get my head around it. The object is to solve for the letter values in the third table.
showimage.php?pid=225207&filename=Picture2.jpg&preview=true
Can someone please help me?

bobbym · 2016-05-17 04:58:57

Hi;

What else do you know about it?

mathisfun_user · 2016-05-17 09:41:44

From A - L, I am guessing that they would be 0, 1, 2, 3, 4, . . . 8.

There's a noticeable pattern when we look at the other two charts and they also go as 0, 1, 2, 3, 4, . . . 8.

Although there are different variables found upon the charts, the third one has to be similar to the other two.

For example, it could be like at one second there is a voltage of 2, at the 2nd second theres a voltage of 19, at the 3rd second theres a voltage of 14 . . .

These charts could represent three different individual testings on the voltages found upon three different electric producing products.

Primarily, I am using this as an example, where these charts could be used in real world application.

Does this help?

Last edited by mathisfun_user (2016-05-17 09:52:39)

thickhead · 2016-05-17 16:30:54

You are right. There is recognizable pattern in the column of 1-digit. 3 digit columns are random and you are not asked to do anything with them.

bobbym · 2016-05-17 16:31:38

Why does the title mention modular arithmetic?

phrontister · 2016-05-19 11:32:17

bobbym wrote:

Why does the title mention modular arithmetic?

In my tables 1, 2 and 3, the second column ('Me') shows the result of modular division on the first column ('3-digit') by the corresponding number in the 'Posn' list, with the results in the second column of tables 1 and 2 giving a reasonable success rate of 13 answers (out of a possible 18) that are the same as those of the OP's. So it seems to me that this modular rule will feature somehow in the solution.

I suppose we need to find a rule that applies to tables 1 and 2 and by which we can solve table 3's letter values...which may not necessarily bear any resemblance to the respective values in tables 1 and 2. Table 3 lists 1 more item than the other tables, but the modular rule caters for that.

      |        1         |        2         |        3
--------------------------------------------------------------
      | 3-digit  1-digit | 3-digit  1-digit | 3-digit  1-digit
Posn  |           Me OP  |           Me OP  |           Me OP
--------------------------------------------------------------
  1   |   856     0  0   |   215     0  0   |   561     0  A
  2   |   176     0  1   |   785     1  1   |   381     1  B
  3   |   751     1  2   |   452     2  2   |   937     1  C
  4   |   343     3  3   |   143     3  3   |   334     2  D
  5   |   652     2  4   |   985     0  4   |   716     1  F
  6   |   977     5  5   |   327     3  5   |   890     2  G
  7   |   447     6  6   |   678     6  6   |   414     1  H
  8   |   583     7  7   |   831     7  7   |   106     2  J
  9   |   296     8  8   |   539     8  8   |   677     2  K
 10   |                                     |   273     3  L

Last edited by phrontister (2016-05-28 01:42:08)

thickhead · 2016-05-23 00:46:03

652 mod 9=985 mod 9=4
176 mod 7=785 mod 7=1
However (751-2) and (452-2) have no common factors and can not be tackled.
Is it of any help?

phrontister · 2016-05-23 02:49:14

Hi thickhead;

I had done something similar when I first looked at this problem and drew up two tables in Excel with 143 columns each (143 being the smallest number seemed to be as good a reason as any to stop there). There are 9 rows, as in the OP's first two tables.

My first table was for identical divisors in each column...which gave a jumbled-looking mess of results.

My second table was the one I mentioned in my previous post (incremented by 1 per row), but stretched out to 143 columns. That gave me the result I posted, which was the one that gave the most hope of containing something meaningful.

Your results appear in my second table.

That table also gave a bit of a sequence in column 6:

{343,143} mod 8 = 7;
{652,985} mod 9 = 4;
{977,327} mod 10 = 7;
{447,678} mod 11 = 7.

But I can't make anything of it, nor of anything else in my two tables.

thickhead · 2016-05-23 04:41:15

We can not change the remainder( 1 digit number given by Patchy 1). we have to accept whatever remainder is given in the table.For {343,143} for the remainder to be 3 we have to find common factor in 343-3 and 143-3 i.e 340 and 140 . The divider may be 5,10 or 20 343 mod 10 or 343 mod 5 etc. All numbers except {751,452} agree.

phrontister · 2016-05-23 13:36:02

Yes, I realise that the remainders (their given figures) are set in concrete...and their given order probably is too.

I wonder, though, if the dividends must also retain their given order. Maybe pairings can be reordered: eg, {752,452} mod 1 then has a common remainder 0. But that approach might upset plans for finding a common rule that can also be applied to table 3.

All numbers except {751,452} agree.

All pairings other than {751,452} have single-digit remainders. That pairing's smallest common remainder (other than for mod 1) is 10, for mod 13.

I can find a listed pairing for each remainder, other than for remainder 5. And it's not possible for all the other remainders to be from different single-digit divisors (though it's not a stated condition), as one of the divisors must be double-digit.

Last edited by phrontister (2016-05-23 15:26:22)

thickhead · 2016-05-23 17:41:29

Also {977,327} does not give proper match.

phrontister · 2016-05-23 17:45:41

This puzzle was posted by patchy1 way back on 8 July 2012, here.

From that thread's first post:

In the 18 examples in the first two boxes below, a simple arithmetic operation on the 3-digit numbers leaves you with the 1-digit numbers. I need to figure out the required simple operation, and translate the ten numbers in the third box into the lettered solutions.
Clues:
A giveaway clue - there can be no 9's in the solution.
The arithmetic operation has two parameters - the given number and another number that is not the same for every case
- and you are looking at it.

Last edited by phrontister (2016-06-03 11:44:29)

phrontister · 2016-05-24 01:22:22

Hi all;

This formula works (it's based on the one given by noelevans in post #2 of the other thread for this puzzle):

Mod(Mod(n,n-r),9)

n = a 3-digit number
r = n's row number (first row = 0)

eg,
Table 1, row 0:
Mod(Mod(856,856-0),9) = 0

Table 2, row 8:
Mod(Mod(539,539-8),9) = 8

Table 3, row 9:
Mod(Mod(273,273-9),9) = 0

My Excel table, with the 1-digit entries being results using my formula:

I had tried to solve it with base 9, but couldn't do it. However, that inspired the Mod 9 addition to noelevans' formula.

Last edited by phrontister (2017-03-31 17:27:27)

Math Is Fun Forum

#1 2016-05-17 01:00:09

Modular arithmetic

#2 2016-05-17 04:58:57

Re: Modular arithmetic

#3 2016-05-17 09:41:44

Re: Modular arithmetic

#4 2016-05-17 16:30:54

Re: Modular arithmetic

#5 2016-05-17 16:31:38

Re: Modular arithmetic

#6 2016-05-19 11:32:17

Re: Modular arithmetic

#7 2016-05-23 00:46:03

Re: Modular arithmetic

#8 2016-05-23 02:49:14

Re: Modular arithmetic

#9 2016-05-23 04:41:15

Re: Modular arithmetic

#10 2016-05-23 13:36:02

Re: Modular arithmetic

#11 2016-05-23 17:41:29

Re: Modular arithmetic

#12 2016-05-23 17:45:41

Re: Modular arithmetic

#13 2016-05-24 01:22:22

Re: Modular arithmetic

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