Find the sum of all positive integers such that their expression in base $7$ digits is the reverse of their expression in base $16$ digits. Express your answer in base $10$.
I tried expressing the digits as $d_1$ $d_2$... and making separate equations but that didn't work. Help!
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If $n=a_k 16^k + \ldots + a_1 16+ a_0= a_0 7 ^k + \ldots a_{k-1}7 + a_k$, with $a_k \geq 1$, then by $n < 7^{k+1}$ we have $16 ^k < 7^{k+1}$. This means that $k\leq 3$ (can manually check this). This limits the search space considerably. Moreover, If the set of coefficients in both expansions is the same, then each $a_i \leq 6$. this makes the problem amenable to a computer search if that suffices for your purposes.
To further reduce the problem, if $a_k = 2$, then $2 \cdot 16^k \leq n < 7^{k+1}$ which gives $k\leq 2$ in this case. This should be fairly easy to code up.
In particular, these numbers have at most as many digits in base seven as they have in base sixteen. So, if $16^{k-1} \leq n < 16^{k}$ is such an integer, then $n$ has at most $k$ digits in base $7$, so $n < 7^k$. Thus $16^{k-1} < 7^k$. This forces $k \leq 3$. By assumption, you know that $n$ has as base-16 digits only $0,1,2,3,4,5,6$.
The one-digit numbers contribute $21$.
For the two-digit numbers: write $n$ as $ab$ in base $16$. Then $16a+b=7b+a$, so $15a=6b$ ie $5a=2b$, so $a=2,b=5$ is the only solution, ie $n=37$.
For the three-digit numbers: write $n$ as $abc$, then $256a+16b+c=49c+7b+a$, thus $255a+9b=48c$, hence $85a+3b=16c$. As $a > 0$, $c > 5$ so $c=6$ and $3|85a$, so $16c=85a+3b > 255$, so $c > 6$, impossible.
So finally it’s $21+37=58$.