# Find a $4$ digit number $abcd$ such that $abcd = 4(dcba)$ [closed]

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 $abcd = cdab$.

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

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• 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 – juantheron Feb 8 '13 at 17:03
• 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. – JB King Feb 8 '13 at 17:07
• Problem 2 is trivial if $a = c$, $b = d$ and $a \neq 0$ – Paresh Feb 8 '13 at 17:07

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...

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.