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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

6 years ago


Answers : (1)


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.

Please feel free to ask as many questions you have.




6 years ago

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