The least number which leaves remainders 2, 3, 4, 5 and 6 on dividing by 3, 4, 5, 6 and 7 is

Question

Problem - The least number which leaves remainders 2, 3, 4, 5 and 6 on dividing by 3, 4, 5, 6 and 7 is?

Solution - Here 3-2 = 1, 4-3 = 1, 5-4 = 1 and so on.

So required number is (LCM of 3, 4, 5, 6, 7) - 1 = 419

My confusion -

I didn't get the solution of this. Please explain how by subtracting the conclusion was drawn that the number will be the (LCM - 1)?

Answer 1

$\, x\!+\!1\equiv 0\pmod{m_i}\iff m_i\mid x\!+\!1\iff {\rm lcm}\{m_i\}\mid x\!+\!1\iff x\equiv -1\pmod{{\rm lcm}\{m_i\}}$

Or, equivalently: $\,\ x\equiv -1\pmod{m_i}\iff x\equiv -1\pmod{{\rm lcm}\{m_i\}},\ $ whihc may be viewed as the special constant-case of the Chinese Remainder Theorem (CRT).

Answer 2

I just found this answer in another similar question

And this explain it.

Thanks.

Answer 3

You can see by inspection that -1 works. Then any other solution must be congruent to -1 mod lcm(3,4,5,6,7) = 420 and hence the least positive solution is 419.