How many two-digit squares have the property that the product of their digits is also a square?
Answers
Answer:
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.
Step-by-step explanation:
Plz mark me as brainliest