a two digit number has different digits. for how many such two digits positive number is the difference between the number itself and the number formed on reversing its digits is the perfect square of an integer ?
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Answers
Answer:
bro this is very difficult question
i know it complex
are you in 11th class???
Step-by-step explanation:
Let the two digits be x and y.
N=10x+y;
Let M be the number obtained on reversing the digits.
M=10y+x;
N+M=10x+y+10y+x=11x+11y=11 (x+y);
N+M is a perfect square; i.e. 11 (x+y) is a perfect square; i.e. 11 divides the perfect square
The maximum value of N+M can be 99+99=198;
Consider the perfect squares series 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196. The next perfect square is 225. We need not go beyond 196 as the maximum value of N+M is 198.
In the above aeries of perfect squares, only 121 is divisible by 11.
Hence M+N=121 i.e. 11 (x+y)=121 i.e. x+y=11 and x, y are singe digit integers.
The integer solutions to the above equation are (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2).
Hence there are eight such two digits numbers such that the sum of the number and its reverse is a perfect square. They are 29, 38, 47, 56, 65, 74, 83, 92.