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# how many 4digit numbers are there with property that it is a square and the number obtained by increasing all its digits by 1 is als a square(kvpy )

Y RAJYALAKSHMI
45 Points
6 years ago
If, each digit of 4 digit number is increased by 1, then the new number is increased by 1111 than the original number.
For eg. consider 1234.  If each digit is increased by 1, then new number is 2345.  The difference of these numbers is 1111.
The number of 4 digit numbers which are perfect squares is 68, ie. from 322 = 1024 to 992 = 9801
Let the original number is of the form a2.
The condition is that, if each digit of these numbers is increased by 1, then the new number also should be a square.  Let the number is of the form b2.  ie. the difference of the new number b2 and the origianl number a2 is 1111.

So, we have b2 – a2 = 1111
=> (b + a) (b – a) = 101* 11
=> b + a = 101 & b – a = 11.
This gives us, b = 56 & a = 45.

Therefore, we have only one number 452 = 2025 when each digit is increased by 1, gives us a perfect square 562 = 3136.

Ans:  only one number. 2025