-1
$\begingroup$

I have two questions:

  1. Find a $4$ digit number $abcd$ such that $abcd = 4(dcba)$.
  2. Find a $4$ digit number $abcd$ such that $abcd = cdab$.
$\endgroup$

closed as off-topic by José Carlos Santos, Leucippus, Cesareo, mrtaurho, Lee David Chung Lin Feb 13 at 11:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Leucippus, Cesareo, mrtaurho, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ actually for (I) i have tried like this way... clearly $a\neq 0$ and $a,b,c,d\in \{0,1,2,3,4,5,6,7,8,9\}$ now If $a=1$ then last digit of $4(dcba)$ is $=4$ and if $a=2$ then last digit of $4(dcba)$ is $ = 8$ So we have got only two values of $a$ i.e $a\in\{1,2\}$now how can i proceed further.. thanks $\endgroup$ – juantheron Feb 8 '13 at 17:03
  • 1
    $\begingroup$ For the second one, isn't a=c,b=d the easy answer? Course there are also a=b=c=d that is also a solution for the second one. $\endgroup$ – JB King Feb 8 '13 at 17:07
  • 2
    $\begingroup$ Problem 2 is trivial if $a = c$, $b = d$ and $a \neq 0$ $\endgroup$ – Paresh Feb 8 '13 at 17:07
2
$\begingroup$

Hints:For 1, $a$ must be even because of the multiplier $4$, so $a=8, d=2$. Now $4*c$ can't carry and $c$ must be odd as it receives a carry of $3$.

For 2, we have $100(ab)+cd=100(cd)+ab$, so $ab=cd$ There are lots...

$\endgroup$
2
$\begingroup$

I'll hint toward the strategy to use for both problems, using $(1)$ as an example:

  1. Find a $4$ digit number $abcd$ such that $abcd = 4(dcba)$.

Note that, taking $abcd$ and $dcba$ to be strings of digits,

We observe that $$abcd = 10^3a + 10^2b + 10 c + d\;\;$$ $$ dcba = 10^3 d + 10^2 c + 10 b + a$$

Now work with the equation: $10^3a + 10^2b + 10 c + d = 4(10^3 d + 10^2 c + 10 b + a)$ and see what you can find to obtain some a, b, c, d satisfying this equation.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.