When a number is divided by $2$ it has quotient $x$ and remainder $1$. Whereas, when the same number is divided by $3$ it has quotient $y$ and remainder $2$. What is the remainder when $x+y$ is divided by $5$?
Problem
Source: Kosovo Math Olympiad 2025, Grade 7, Problem 4
Tags: remainder, number theory, Kosovo, national olympiad
18.11.2024 00:00
We have $n=2x+1$ and $n=3y+2$. Hence, $$2x+1=3y+2 \Rightarrow 2x-3y=1.$$We know $-3y\equiv 2y \pmod 5$, so $2x+2y \equiv 1 \pmod 5$. Checking all the cases, we conclude $\boxed{x+y \equiv 3 \pmod 5}$.
18.11.2024 09:49
Quote: Checking all the cases, we conclude $\boxed{x+y \equiv 3 \pmod 5}$. Alternatively, multiply by $2^{-1}$ on both sides.
18.11.2024 14:02
AshAuktober wrote: Quote: Checking all the cases, we conclude $\boxed{x+y \equiv 3 \pmod 5}$. Alternatively, multiply by $2^{-1}$ on both sides. Yeah, that's essentially what I did (use the fact that $5$ is odd), but remember this is an olympiad for 7th grade
18.11.2024 15:35
This stuff is probably more suited for HSM based on the difficulty. Also does this comp require proofs or is it short answer? If the latter, then one could simply notice that $\tfrac{5}{2}=2R1$ and $\tfrac{5}{3}=1R2,$(where $\tfrac{a}{b}=cRd$ if $a=bc+d$ and $d<b$, so $c$ is the quotient and $d$ is the remainder) so the answer is $1+2=\boxed{3}.$
24.11.2024 19:09
Let a be a number such that a=2x+1=3y+2. We have that x+y=(5a-7) : 6, so 6(x+y)=5a-7, where we get that x+y is 3 (mod 5).