Find the four smallest four-digit numbers that meet the following condition: by dividing by $2$, $3$, $4$, $5$ or $6$ the remainder is $ 1$.
Problem
Source:
Tags: number theory, remainder
Arr0w
13.09.2021 21:33
We have \begin{align*} x&\equiv 1 \pmod 2\\ x&\equiv 1 \pmod 3\\ x&\equiv 1 \pmod 4\\ x&\equiv 1 \pmod 5\\ x&\equiv 1 \pmod 6.\\ \end{align*}Applying the Chinese Remainder Theorem we find that $x\equiv 1\pmod{60}$ or that $x=60n+1$ for some natural $n$. The problem asked for the smallest first four digit numbers that match this condition, and by guess and check we get that $n\ge 17$ so our final answer $x=\boxed{1021,1081, 1141, 1201}$.
OlympusHero
14.09.2021 05:06
Let such a number be $x$. Then $x \equiv 1 \pmod {60}$, since $60$ is the least common multiple of $2,3,4,5,6$. Our smallest possible values are hence $\boxed{1021, 1081, 1141, 1201}$.
Saucepan_man02
14.09.2021 08:39
Let the number be $ N $, then $ N = (lcm(2,3,4,5,6))k + 1 = 60k+1 $, where k is a positive integer. Now, $ N $ is the 4 least 4-digit number of the form $ 60k+1 $. After a bit of Errorless Working , we get $ N = 1021, 1081, 1141, 1201 $