WakeUp wrote: Find all the integers written as $\overline{abcd}$ in decimal representation and $\overline{dcba}$ in base $7$. $(2116)_{10}=(6112)_{7}$ so result is $\boxed{2116}$

roza2010 wrote: WakeUp wrote: Find all the integers written as $\overline{abcd}$ in decimal representation and $\overline{dcba}$ in base $7$. $(2116)_{10}=(6112)_{7}$ so result is $\boxed{2116}$ $\overline {0000}_{10}$ is also a solution

pco wrote: roza2010 wrote: WakeUp wrote: Find all the integers written as $\overline{abcd}$ in decimal representation and $\overline{dcba}$ in base $7$. $(2116)_{10}=(6112)_{7}$ so result is $\boxed{2116}$ $\overline {0000}_{10}$ is also a solution Quite right roza2010, you should perhaps elaborate even though it is quite a simple problem.

I know the basis of this exercise and i have gone this far: 333a+31b-13c-114d=0 How do I develop the exercise further?

$\overline {abcd}_{10}$ = $\overline {dcba}_{7}$ $\implies$ $1000a + 100b + 10c + d = 343d + 49c + 7b + a \implies\ 999a + 93b = 342d + 39c$ or $333a + 31b = 114d + 13c$. If $a = 3$ $\implies$ $999 + 31b = 114d + 13c$. If we put d and c maximum we get that $999 + 31b = 762$,impossible $b \ge\ 0$ $\implies a < 3$. If $a = 0$ $\implies 31b = 114d + 13c$,if $d = 2$ $\implies 31b = 228 + 13c$,impossible $\implies d < 2$. If $d = 1$ there aren't solutions,so the solution is for $d = 0$,$c = 0$ and $b = 0$. If we analyse cases where $a = 1$ or $a = 2$ we get another solution:$b = c = 1,d = 6$. The solutions are $\overline {abcd}_{10}$ = {$\overline {0000}_{10}$,$\overline {2116}_{10}$}.

We have $1000a+100b+10c+d = 343d+49c+7b+a$, or $999a+93b = 39c + 342d \implies 333a+31b = 13c + 114d$. Note that the maximum value of the RHS is $1143$, less than $333 \cdot 4$, so we only test $a = 1, 2, 3$. If $a = 1$, then $d = 2$, but this gives no solutions. If $a = 2$, then $d=5$ gives no solutions, but $d = 6$ gives $\boxed{2116}$ in base $10$. Similarly, $a = 3$ gives no solutions.

1000a+100b+10c+d=343d+49c+7b+a, or 333a+31b=114d+13c. Even if d=c=9, a can only go up to 3, so we can casework on them. If a=1, then d is 2, if d is 3 RHS too large if d=1 then 114+13x9 < 333. However, 105=13c-31b gives no solutions, one way to check is mod 13 meaning -31b=8b is 1 mod 13, and testing (since 8x9 isn't that time consuming) we have no solutions. Similarly for a=2, then d=5 no solutions but d=6 gives the solution $\boxed{2116}$. (d=7 also has no sols.) Now also note that 0 is a solution, assuming that a can be 0. Very bashy problem.. but ig because it is shortlist

We gave that $1000a+100b+10c+d=343d+49c+7b+a$, so $333a+31b=13c+114d$, note that the only values of $a$ can be $1,2,3$ so we test, if $a=1$ and $a=3$ give no solutions, if $a=2$ we have $2116$ works.