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## Smallest number

Posted by Chris on June 1, 2011 – 2:44 pm

What is the smallest number that when you divide it by 2,3,4,5 and 6 will have remainders 1, 2, 3, 4 and 5 respectively?

This post is under “MathsChallenge” and has 8 respond so far.

### 8 Responds so far- Add one»

1. 1. Karys Said：

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.

2. 2. Mody Said：

it’s 59

3. 3. Michelle Said：

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

4. 4. Michelle Said：

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

5. 5. Chris Said：

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

6. 6. anastacio Said：

7. 7. Chris Said：

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

8. 8. prashANT Said：

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.

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