If \(x\) is a positive integer, the value of the function \(\mathrm{d}(x)\) is the sum of the digits of \(x\) in base 10. For example, \(\mathrm{d}(249) = 2 + 4 + 9 = 15\).
An \(n\)-digit positive integer \(x\) is written in the form \(\displaystyle\sum_{r=0}^{n-1} a_r \times 10^r\), where \(0 \leqslant a_r \leqslant 9\) for all \(0 \leqslant r \leqslant n-1\) and \(a_{n-1} > 0\).
- Prove that \(x - \mathrm{d}(x)\) is non-negative and divisible by \(9\).
- Prove that \(x - 44\mathrm{d}(x)\) is a multiple of \(9\) if and only if \(x\) is a multiple of \(9\).
Suppose that \(x = 44\mathrm{d}(x)\). Show that if \(x\) has \(n\) digits, then \(x \leqslant 396n\) and \(x \geqslant 10^{n-1}\), and hence that \(n \leqslant 4\).
Find a value of \(x\) for which \(x = 44\mathrm{d}(x)\). Show that there are no further values of \(x\) satisfying this equation.
- Find a value of \(x\) for which \(x = 107\mathrm{d}\left(\mathrm{d}(x)\right)\). Show that there are no further values of \(x\) satisfying this equation.
The integers \(a,b\) and \(c\) satisfy
\[
2a^{2}+b^{2}=5c^{2}.
\]
By considering the possible values of \(a\pmod5\) and \(b\pmod5\), show
that \(a\) and \(b\) must both be divisible by \(5\).
By considering how many times \(a,b\) and \(c\) can be divided by \(5\),
show that the only solution is \(a=b=c=0.\)
Show Solution
\begin{array}{c|ccccc}
a & 0 & 1 & 2 & 3 & 4 \\
a^2 & 0 & 1 & 4 & 4 & 1
\end{array}
Therefore \(a^2 \in \{0,1,4\}\) and so we can have
\begin{array}
$2a^2+b^2 & 0 & 1 & 4 \\ \hline
0 & 0 & 1 & 4 \\
1 & 2 & 3 & 1 \\
4 & 3 & 4 & 2
\end{array}
Therefore the only solution must have \(5 \mid a,b\), but then we can write them has \(5a'\) and \(5b'\) so the equation becomes
\(2\cdot25 a'^2 + 25b'^2 = 5c^2\) ie \(5 \mid c^2 \Rightarrow 5 \mid c\). But that means we can always divide \((a,b,c)\) by \(5\), which is clearly a contradiction if we consider the lowest power of \(5\) dividing \(a,b,c\) for any solution.
A secret message consists of the numbers \(1,3,7,23,24,37,39,43,43,43,45,47\) arranged in some order as \(a_{1},a_{2},\ldots,a_{12}.\) The message is encoded as \(b_{1},b_{2},\ldots,b_{12}\) with \(0\leqslant b_{j}\leqslant49\) and
\begin{alignat*}{1}
b_{2j} & \equiv a_{2j}+n_{0}+j\pmod{50},\\
b_{2j+1} & \equiv a_{2j+1}+n_{1}+j\pmod{50},
\end{alignat*}
for some integers \(n_{0}\) and \(n_{1}.\) If the coded message is \(35,27,2,36,15,35,8,40,40,37,24,48,\)
find the original message, explaining your method carefully.
Show Solution
Considering the odd numbers, we have
\begin{array}{l|rrrrrr}
b_{2j+1} & 35 & 2 & 15 & 8 & 40 & 24 \\
a_{2j+1}+n_1 & 35 & 1 & 13 & 5 & 36 & 19
\end{array}
Considering the even numbers, we have
\begin{array}{l|rrrrrr}
b_{2j} & 27& 36 & 35 & 40 & 37 & 48 \\
a_{2j}+n_0 & 27 & 35 & 33 & 37 & 33 & 43
\end{array}
There are three numbers in the original sequence which are repeated (\(43\)). By the pigeonhole principle, one of the odds or evens must have at least two of them. We can see that the even numbers have some number repeated twice (\(33\)). Therefore these must be the \(43\)s. Therefore \(n_0 = -10\)
\begin{array}{l|rrrrrr}
b_{2j} & 27& 36 & 35 & 40 & 37 & 48 \\
a_{2j}+n_0 & 27 & 35 & 33 & 37 & 33 & 43 \\
a_{2j} & 37 & 45 & 43 & 47 & 43 & 3
\end{array}
This leaves the remaining numbers to be decoded from the original sequence as \(1,7,23,24,39,43\). Two of these numbers are consecutive (\(23\) and \(24\)), and two numbers in our sequence are \(35\) and \(36\). Therefore \(n_1\) must be \(12\).
\begin{array}{l|rrrrrr}
b_{2j+1} & 35 & 2 & 15 & 8 & 40 & 24 \\
a_{2j+1}+n_1 & 35 & 1 & 13 & 5 & 36 & 19 \\
a_{2j+1} & 23 & 39 & 1 &43 & 24& 7
\end{array}
Therefore the original sequence was: \(23, 37, 39, 45, 1, 43, 43, 47, 24, 43, 7, 3\)