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```        find a 4 digit number M such that number N = 4M satisfies the following properties -
a) N is also a four digit number
b) N has the same digits as M but in reverse order```
7 years ago

22 Points
```										Hi Mihir,
The Answer is M = 2178 and N = 8712.
You can find the values by proceeding in follwing way,
M can lie between 1000 and 2499 and N can lie between 4000 and 9996 as both are 4 digit Number and N = 4M
Now since N and M have reversed digits so,
M has to be between 2000 and 2499 becasue N is even(N=4M) ( N can't end with 1, it should end with 2 )
Now we have starting digit as 2 for M and ending as 2 for N.
Now N = 4M so , N has to start with 8 or 9.
Considering the first possibilty of N starting with 8 ,
When N starts with 8 then M ends with 8 can be only 1 ,2 ,3 ,4 ( M lies between 2000 and 2499 )
so third digit of N shud be 1,2,3 or 4
Now by multiplying 8 by 4 we get 32 i.e 2 remains in end and 3 is carried forward.
Since 3 is carried forward so 3 digit of N shud be odd ( because 3 + 4(any digit) = odd )
Hence 1 or 3 as the choice remains for 3rd digit of N
by putting as 1 , u will get N as 8(blank)12 and M as 21(blank)8
By making more INtelligent guesses in this manner u can get the number as N = 8712 and 2178
( because 7 or 2 gives 28 or 8 which on addition of 3 will give 31 or 11 ending in 1 )
I hope I am clear with the answer.
Puneet

```
7 years ago
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