Try this problem of TIFR GS-2010 using your concepts of number theory and congruence based on Positive Integers.
If $n$ and $m$ are positive integers and $n^{19}=19m+r$ then the possible values for $r$ modulo $19$ are:-
NUMBER THEORY
CONGRUENCE
FERMAT'S LITTLE THEOREM
Answer:only $0$,$\pm1$
TIFR 2010|PART B |PROBLEM 12
ELEMENTARY NUMBER THEORY DAVID M.BURTON
$r(mod 19)=(n^9-19m)(mod 19)=n^9(mod 19)-19m(mod 19)=n^9(mod 19)$ [because $19\mid19m$]
Now there can be two cases
casei) $19\mid n$
caseii) n is not divisible by 19
If $19\mid n$ then $19\mid n^9$ resulting in $n^9\equiv 0(mod 19)$
But if $n$ is not divisible by $19$ then according to Fermat's little theorem,
$n^{19-1}\equiv1(mod 19) =n^{18}\equiv 1(mod 19) = 19\mid n^{18}-1 =19\mid[n^9+1][n^9-1]$
i.e If $19\mid n^9+1$ then $n^9\equiv(-1)(mod 19) =r\equiv(-1)(mod 19)$
If $19\mid n^9-1$ then $n^9\equiv1(mod 19) = r\equiv1(mod 19)$
Thus $r(mod 19)=0$,$\pm1$
In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]
Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.
Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.
Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.