With modulos it becomes : Find the smallest integer n such that
n = 1 (2)
n = 2 (3)
n = 3 (4)
n = 4 (5)
n = 5 (6)

Redundant equations :
n = 1 (2) is already implied by n = 3 (4) and n = 5 (6) (it just means that n is odd).

n = 1 (2) is also equivalent to :
n = 1 OR 3 OR 5 (6)

n = 2 (3) is :
n = 2 OR 5 (6)

so we end up with only 3 equations :
n = 3 (4)
n = 4 (5)
n = 5 (6)

you can also write from n = 3 (4) :
n = 3 OR 7 OR 11 OR 15 OR 19 (20)
and from n = 4 (5)
n = 4 OR 9 OR 14 OR 19 (20)

so we get n = 5 (6) and n = 19 (20)

same thing at 60 :
n = 5 (6) is n = 5 OR 11 OR 17 OR 23 OR 29 OR 35 OR 41 OR 47 OR 53 OR 59 (60)
n = 19 (20) is n = 19 OR 39 OR 59 (60)

so we end up with 59.

I admit that may have been too quick an explanation, I mainly did not bother to explain how I “expanded” the equations to get equations with the same modulo.

Hi Michelle, I think you must be jumbling something up.
59/2 = 29 r 1, 59/3 = 19 r 2, 59/4 = 14 r 3, 59/5 = 11 r 4 and 59/6 = 9 r 5.

— SOLUTION
Let n be the number sought after. For each case we have n = m – 1 (mod m).
Adding 1 to both sides gives: n + 1 = m = 0 (mod m), where m = 2,3,4,5,6

So we need n + 1 = lcm(2,3,4,5,6) = 60, so n = 60 – 1 = 59.

To get the lcm, note that the only prime factors for 2-6 are 2,3,5.
But 4 = 2², so the lcm is 2²*3*5 = 60

Hi anastacio. Nice try, 11 nearly works. It only fails for the divide by 5 case, but then you get a remainder 1 or -4.

I also note that if I’d extended the original problem, to an arbiitrary extent, e.g. n = 99 (mod 100), then n would have been lcm(2,3,4,5,…,100) – 1 = 69720375229712477164533808935312303556799

ITS 59 AS-:
IT LEAVES 1 AS REMAINDER WHEN DIVIDED BY 2 .SO ITS AN ODD NO.
IT LEAVES 2 AS REMAINDER WHEN DIVIDED BY 3 . SO IT CAN BE 5,11,17 AND SO ON
IT LEAVES 3 AS REMAINDER WHEN DIVIDED BY 4 . SO IT CAN BE 11,23,35,47,59,71 AND SO ON
IT LEAVES 4 AS REMAINDER WHEN DIVIDED BY 5 SO IT CAN END IN 4 OR 9 BUT ITS AN ODD NO SO IT ENDS IN 9.IN ABOVE SERIES 59 IS THE SMALLEST NO AND IT ALSO GIVES 5 AS REMAINDER WHEN DIVIDED BY 6.

PHP Warning: PHP Startup: Unable to load dynamic library 'C:\Program Files (x86)\Parallels\Plesk\Additional\PleskPHP5\ext\php_mssql.dll' - The specified module could not be found.
in Unknown on line 0
PHP Warning: PHP Startup: Unable to load dynamic library 'C:\Program Files (x86)\Parallels\Plesk\Additional\PleskPHP5\ext\php_pdo_mssql.dll' - The specified module could not be found.
in Unknown on line 0

June 1st, 2011 at 3:14 pm

With modulos it becomes : Find the smallest integer n such that

n = 1 (2)

n = 2 (3)

n = 3 (4)

n = 4 (5)

n = 5 (6)

Redundant equations :

n = 1 (2) is already implied by n = 3 (4) and n = 5 (6) (it just means that n is odd).

n = 1 (2) is also equivalent to :

n = 1 OR 3 OR 5 (6)

n = 2 (3) is :

n = 2 OR 5 (6)

so we end up with only 3 equations :

n = 3 (4)

n = 4 (5)

n = 5 (6)

you can also write from n = 3 (4) :

n = 3 OR 7 OR 11 OR 15 OR 19 (20)

and from n = 4 (5)

n = 4 OR 9 OR 14 OR 19 (20)

so we get n = 5 (6) and n = 19 (20)

same thing at 60 :

n = 5 (6) is n = 5 OR 11 OR 17 OR 23 OR 29 OR 35 OR 41 OR 47 OR 53 OR 59 (60)

n = 19 (20) is n = 19 OR 39 OR 59 (60)

so we end up with 59.

I admit that may have been too quick an explanation, I mainly did not bother to explain how I “expanded” the equations to get equations with the same modulo.

June 1st, 2011 at 3:36 pm

it’s 59

June 1st, 2011 at 5:00 pm

If you go with 59 as the number than you don’t get the remainders of 1, 2, 3, 4, and 5! Explain that!!

June 1st, 2011 at 5:01 pm

I thought it would be 59 as well, but the remainders are not what the riddle is looking for!!

June 1st, 2011 at 7:23 pm

Hi Karys, you got it right.

Hi Michelle, I think you must be jumbling something up.

59/2 = 29 r 1, 59/3 = 19 r 2, 59/4 = 14 r 3, 59/5 = 11 r 4 and 59/6 = 9 r 5.

— SOLUTION

Let n be the number sought after. For each case we have n = m – 1 (mod m).

Adding 1 to both sides gives: n + 1 = m = 0 (mod m), where m = 2,3,4,5,6

So we need n + 1 = lcm(2,3,4,5,6) = 60, so n = 60 – 1 = 59.

To get the lcm, note that the only prime factors for 2-6 are 2,3,5.

But 4 = 2², so the lcm is 2²*3*5 = 60

June 2nd, 2011 at 11:49 pm

the answer is 11

June 3rd, 2011 at 3:47 am

Hi anastacio. Nice try, 11 nearly works. It only fails for the divide by 5 case, but then you get a remainder 1 or -4.

I also note that if I’d extended the original problem, to an arbiitrary extent, e.g. n = 99 (mod 100), then n would have been lcm(2,3,4,5,…,100) – 1 = 69720375229712477164533808935312303556799

June 6th, 2011 at 2:54 am

ITS 59 AS-:

IT LEAVES 1 AS REMAINDER WHEN DIVIDED BY 2 .SO ITS AN ODD NO.

IT LEAVES 2 AS REMAINDER WHEN DIVIDED BY 3 . SO IT CAN BE 5,11,17 AND SO ON

IT LEAVES 3 AS REMAINDER WHEN DIVIDED BY 4 . SO IT CAN BE 11,23,35,47,59,71 AND SO ON

IT LEAVES 4 AS REMAINDER WHEN DIVIDED BY 5 SO IT CAN END IN 4 OR 9 BUT ITS AN ODD NO SO IT ENDS IN 9.IN ABOVE SERIES 59 IS THE SMALLEST NO AND IT ALSO GIVES 5 AS REMAINDER WHEN DIVIDED BY 6.