What's the smallest number that's divisible by all numbers 1 thru 20 without leaving a remainder?

[Yavin4]

???

[bluedigger]

1. I understood there would be no math questions in the Lounge.

[pinboy3niner]

22. There's no math in the Lounge!1!!

[Xipe Totec]

2. product(prime(range(1...20))) nt

What I meant was:

1*2*3*5*7*11*13*17*19

or

9,699,690

[Yavin4]

5. Incorrect. n/t

[hunter]

6. Sadly, it fails at four.

9,699,690/4 = 2424922.5

[Xipe Totec]

7. Forget accuracy, look at the speed!

[caraher]

11. Ah, the Max Power way!!!

From The Simpsons... see following post for link that works! Mine was done the Max Power way!

[hunter]

12. Google can fix that too!

[Xipe Totec]

14. That's it! Exactly!

[Downwinder]

8. 9 699 690 / 20 = 484 984.5

[Xipe Totec]

9. You want speed, or do you want accuracy?

[Scuba]

3. Execute Plan Blue. Repeat. Execute Plan Blue.

[hunter]

4. Answer: Google is awesome.

Spoiler Warning:

http://codeblog.dhananjaynene.com/2010/01/least-common-multiple-what-is-the-smallest-number-that-is-evenly-divisible-by-all-of-the-numbers-from-1-to-20

My first thought followed the Julian/Mathew fellow. I knew I'd be throwing the primes around starting with 19.

[rug]

10. 87?

[fishwax]

13. I believe it is

232,792,560 ...

On edit--that can't be right. Maybe 19.399,380.

[Yavin4]

15. Your First Answer is Correct.

You are the smartest kid in class.

[fishwax]

17. d'oh! that's what I get for changing my answer

[Tuesday Afternoon]

16. show your work to the class, please

[Art_from_Ark]

18. Calculation

5 X 7 X 9 X 11 X 13 X 16 X 17 X 19 = 232,792,560

Explanation:
Must include all prime numbers (2,3,5,7,11,13,17,19), plus must be divisible by 9 (3 X 3) and 16 (2 X 2 X 2 X 2), so the prime numbers 2 and 3 by themselves can be excluded from the calculation.

[eppur_se_muova]

19. 2^4*3^2*5*7*11*13*17*19

i.e. the product of all prime powers <= 20

Method works for any number.

[Yavin4]

21. Incorrect.

The product of all prime numbers <=20 is 9699690, which is incorrect.

[eppur_se_muova]

26. No, correct. Read carefully -- all prime *powers*. That's why I wrote 2^4, not 2, etc.

This certainly works, and has useful applications in computational number theory -- see, for example http://en.wikipedia.org/wiki/Pollard's_p_%E2%88%92_1_algorithm#Algorithm_and_running_time . (Step 2 takes the product of all prime powers <= B.)

[Yavin4]

29. I Stand Corrected

I mis-read your post.

[eppur_se_muova]

30. Always blame your bad glasses prescription.

On the Internet, no one knows if you wear glasses or not.

[hkis5269se]

20. Uhhhhh... this is hard.

[Liberal Veteran]

23. Is it the same as the airspeed velocity of the unladen sparrow?

I'm going to go with 42 since it my go to answer for everything.

[shadowrider]

31. African or European?

[Dr. Strange]

24. In other words...

what's the least common multiple of the numbers 1 through 20.

Focusing on the primes, what is the highest power of each prime which is less than or equal to 20?
Quick formula: for a prime p, use FLOOR(log(20)/log(p))

p=2: t=FLOOR(log(20)/log(2)) = 4
(Or you could just say, hey, 2⁴ <= 20 < 2⁵.)

p=3: t=FLOOR(log(20)/log(3)) = 2

p=5 through 19: t=FLOOR(log(20)/log(p)) = 1

So the answer is 2⁴3²5¹1¹11¹13¹17¹19¹ = 232792560.

[Dr. Strange]

25. EXTRA CREDIT

Let N denote the smallest positive integer which is divisible by all integers from 1 to 100. Find the smallest positive integer, in terms of N, which is divisible by all integers from 1 to 105.

[eppur_se_muova]

28. 101*103*N nt

[Dr. Strange]

32. Winner!

You get five extra minutes in the Lounge hot tub!

[Ikonoklast]

27. You didn't say this was gonna be on the test!

All week, we asked you, "Is this gonna be on the test? Is this gonna be on the test?"

You never said, "YES, this will be on the test!"

NOW IT'S ON THE TEST!!!

I'm telling Principal Skinner!!