Juca has decided to call all positive integers with 8 digits as $sextalternados$ if it is a multiple of 30 and its consecutive digits have different parity. At the same time, Carlos decided to classify all $sextalternados$ that are multiples of 12 as $super sextalternados$. a) Show that $super sextalternados$ numbers don't exist. b) Find the smallest $sextalternado$ number.
Problem
Source: Lusophon Matematical Olympiad 2021 Problem 1
Tags: number theory
StarLex1
20.12.2021 05:42
Part B $(abcdefgh)$ different parity mean = oddevenoddeven and so on or evenoddevenodd and so on divisible by 5 h must be = $(0,5)$ divisible by 2 h must be = $(0,2,4,6,8)$ divisible by 3 must be = $3|(a+b+c+d+e+f+g+h)$ h that is possible $h = 0$ take $b,d,f =0 $ $3|(a+b+c+d+e+f+g+0)$ $3|(a+c+e+g) =(1+1+1+3) $ take g the biggest among a,c,e,g because g have lowest value so the smallest sextalternado number $(10101030)$ Part A $lcm (30,12) = 60$ $60 = 2^2*3*5$ divisible by 4 = $4|(gh)$ gh is decimal representation not multiplication divisible by 5 = $(0,5) $ divisible by 3 same as above so the possible value of h is 0 since 4 is even if h is even g must be odd 10,30,50,70,90 none of them is divisible by 4
Jalil_Huseynov
20.12.2021 21:30
a) $super sextalternados$ number is divisible by $5$ and it's even. So last number of it os $0$. Since second digit from end should be odd, it can give $10,30,50,70,90$ as a residue when divided by $100$,but then it's not divisible by $4$. b) Obviously $10101030$ works. Assume there exists less $sextalternados$ than it and let it be $t$ . Then obviously $t=101010a0$ and since $t<10101030$, we get $t=10101010$. But then $3\nmid t$. So $10101030$ is the smallest one.