Finding the least dividend

Question

Find the least number when divided by $\ 2,3,4,5,6$ leaves the remainder of $\ 1 $ in each case but when divided by $ 7$ leaves no remainder. Is there a way of doing this other than hit and trail and if no then what is the standard way of using hit and trial in these types of questions

When the numbers are small enough, it might not be too bad to run through the Chinese remainder algorithm by hand. But you can have Wolfram Mathematica or Wolfram Alpha compute it for you: ChineseRemainder[{1, 1, 1, 0}, {4, 5, 3, 7}]. See Peter's answer and http://reference.wolfram.com/language/ref/ChineseRemainder.html — Robert Soupe, May 31 '15 at 18:37

score 5 · Accepted Answer · answered May 31 '15 at 17:56

Suppose that $n$ has the desired properties (apart from minimality). Then $n-1$ is divisible by $2,3,4,5$, and $6$, so it must be divisible by $60$, the least common multiple of these numbers. Thus, you’re looking for a number of the form $60k+1$ that is divisible by $7$. $60\equiv4\pmod7$, so $60k+1\equiv4k+1\pmod7$, and it’s easy now to see by inspection that $k=5$ is the minimum positive solution. Thus, $n=60\cdot5+1=301=7\cdot43$.

score 3 · Answer 2 · edited May 31 '15 at 18:32

3

Hint: Use the Chinese remainder theorem.

It is sufficient that

$$x \equiv 1 \pmod 4$$ $$x \equiv 1 \pmod 5$$ $$x \equiv 1 \pmod 3$$ $$x \equiv 0 \pmod 7$$

The general solution is $x = 420k + 301$.

edited May 31 '15 at 18:32

Robert Soupe

14,999

answered May 31 '15 at 17:56

Peter

86,576

Finding the least dividend

2 Answers2