Remainder of 1

[Home]   [Puzzles & Projects]    [Delphi Techniques]   [Math topics]   [Library]   [Utilities]

 

Search

Search WWW

Search DelphiForFun.org

As of October, 2016, Embarcadero is offering a free release of Delphi (Delphi 10.1 Berlin Starter Edition ).     There are a few restrictions, but it is a welcome step toward making more programmers aware of the joys of Delphi.  They do say "Offer may be withdrawn at any time", so don't delay if you want to check it out.  Please use the feedback link to let me know if the link stops working.

 

Support DFF - Shop

 If you shop at Amazon anyway,  consider using this link. 

     

We receive a few cents from each purchase.  Thanks

 


Support DFF - Donate

 If you benefit from the website,  in terms of knowledge, entertainment value, or something otherwise useful, consider making a donation via PayPal  to help defray the costs.  (No PayPal account necessary to donate via credit card.)  Transaction is secure.

Mensa Daily Puzzlers

For over 15 years Mensa Page-A-Day calendars have provided several puzzles a year for my programming pleasure.  Coding "solvers" is most fun, but many programs also allow user solving, convenient for "fill in the blanks" type.  Below are Amazon  links to the two most recent years.

Mensa 365 Puzzlers  Calendar 2017

Mensa 365 Puzzlers Calendar 2018

(Hint: If you can wait, current year calendars are usually on sale in January.)

Contact

Feedback:  Send an e-mail with your comments about this program (or anything else).

Search DelphiForFun.org only

 

 

Problem Description

What is the smallest multiple of 13 which leaves a remainder of 1 when
divided by each of the numbers 2 through 12?

Background & Techniques

We'll try 2 methods to find a solution:

bulletBrute force- testing successive multiples of 13 until we find one
that leaves a remainder of 1 when divided by 2 through 12.
bulletUsing LCM - find lowest common multiple (LCM) of 2 through 12.  This LCM, or any multiple of this LCM,  plus 1 must leave a remainder of 1 when divided by 2 through 12. We'll just start testing successive multiples of LCM + 1 until we find one that is divisible by 13.

Process vs. Product

Like many of the problems here on DFF, the answer here  is less significant  than the solving process.  Because the code is so simple, I've beefed it up just a little by adding the second solution technique.  It provides a good chance to review Lowest  Common Multiple (LCM) and it's relationship to Greatest Common Denominator (GCD).     Given two integers, a and  b, it is always the case that GCD(a,b)LCM(a,b)=ab.   There a GCD page over in the Math Topics sect with more information about GCD.  

By the way, I learned  Lowest Common Multiple, but current usage seems to favor Least Common Multiple by about 3 to 1.  Google finds 2,100,000 "Least  Common Multiple" references and only 710,000  for "Lowest  Common Multiple"!!  

One more "freebie" - it occurred to me that, aside from the educational benefits of using the LCM method, it should be much faster.  About half a dozen extra lines of code, tells us the answer - 70 times faster  (636 vs. 9 microseconds on my Celeron 800mhz system.)  Windows procedures QueryPerformanceCounter and QueryPerofrmanceFrequency make accurate timing pretty simple.  Now if I can just make wise use of the time saved!

Running/Exploring the Program 

bulletBrowse source extract
bulletDownload source
bulletDownload  executable

Suggestions for Further Explorations

???

 
Original Date: August 24, 2002 

Modified: July 29, 2017

 
  [Feedback]   [Newsletters (subscribe/view)] [About me]
Copyright 2000-2017, Gary Darby    All rights reserved.