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### 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:

 Brute force- testing successive multiples of 13 until we find one that leaves a remainder of 1 when divided by 2 through 12. Using 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)=a×b.   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!

### Suggestions for Further Explorations

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 Original Date: August 24, 2002 Modified: July 29, 2017