Question

Let n = 3^17 + 3^10 It is known that 11 divides into n+1. If n can be written in base 10 as ABCACCBAB, where A,B,C are distinct digits such that A and C are odd and B is not divisible by 3, find 100A + 10B + C.

Answer 1

first find \(A\) and \(B\) from
\(\large \color{black}{\begin{align} n\pmod {100}\hspace{.33em}\\~\\
\end{align}}\)
ABCACCBAB
then use the divisiblity criteria of \(11\) that
\(\large \color{black}{\begin{align} B+A+C+A-(A+C+C+B+(B+1))=11n\hspace{.33em}\\~\\
\end{align}}\)

where \(n=0,1,2,\cdots ,+\infty\)

Answer 2

does this makes sense

Answer 3

:-) not much Sir, trying out other things

Answer 4

difficulty in \(n \pmod {100}\) ?

Answer 5

yes

Answer 6

\(\large \color{black}{
\dfrac{3^{17} + 3^{10}}{100}\hspace{.33em}\\~\\
\implies \dfrac{3^{10}\left(3^{17} + 1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{9^{5}\left(3^{7} + 1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{(10-1)^{5}\left(3^{7} + 1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{\binom{5}{0}\cdot (-1)^5\cdot(10)^0+\binom{5}{1}\cdot (-1)^4\cdot(10)^1+\cdots )\left(3^{7} + 1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{(-1+50 )\left(3^{7} + 1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{(49 )\left(3^{7} + 1\right)}{100}\hspace{.33em}\\~\\
\color{red}{\implies}\dfrac{3^{7}}{100}\hspace{.33em}\\~\\
\color{red}{\implies}\dfrac{3\cdot 3^{6}}{100}\hspace{.33em}\\~\\
\color{red}{\implies}\dfrac{3\cdot 9^{3}}{100}\hspace{.33em}\\~\\
\color{red}{\implies}\dfrac{3\cdot 729}{100}\hspace{.33em}\\~\\
\color{red}{\implies}\dfrac{3\cdot 29}{100}\hspace{.33em}\\~\\
\color{red}{\implies}\dfrac{87}{100}\hspace{.33em}\\~\\
\color{red}{\implies}\dfrac{-13}{100}\hspace{.33em}\\~\\
\implies \dfrac{(49 )\left(-13 + 1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{-588}{100}\hspace{.33em}\\~\\
\implies \dfrac{-88}{100}\hspace{.33em}\\~\\
\implies \dfrac{12}{100}\hspace{.33em}\\~\\
\LARGE \implies 12\hspace{.33em}\\~\\
A=1,B=2\hspace{.33em}\\~\\
}\)

Answer 7

\(\large \color{black}{ B+A+C+A-(A+C+C+B+(B+1))=11n\hspace{.33em}\\~\\
2A+B+C-(A+2C+2B+1)=11n\hspace{.33em}\\~\\
A-B-C=11n+1\hspace{.33em}\\~\\
1-2-C=11n+1\hspace{.33em}\\~\\
C=-11n-2\hspace{.33em}\\~\\
\normalsize \text{u want C as single digit positive number so}\hspace{.33em}\\~\\
C=-11(-1)-2\hspace{.33em}\\~\\
C=9\hspace{.33em}\\~\\
}\)

Answer 8

A=1,B=2 and C=9.