Three digit number
Algebra
Level
pending
Find a three digit number, such that- 1)Every digit is nonzero, 2)The sum of the square of its first two digits is equal to the square of its last digit, 3)The first digit is prime.
The answer is 345.
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1 solution
Let, the digits of the numbers are x,y,z. Now, according to the problem, x^2+y^2 =z^2
or, x^2 = z^2 - y^2 or, x^2 = (z+y)(z-y) Now , x is prime. Then we get two possibilities from this condition - 1) x = (y+z) , and x = (y-z) , implies that z = 0,which is not possible , because every digit is nonzero. hence, the other possibility is, x^2 = (z+y) ,and 1 = (z-y). then,( x^2 - 1 ) = 2y or, (x+1)(x-1) = 2y implies that , if (x-1) = 2, then (x+1) = y . hence, we get x = 3 and y = 3+1 =4, implies that z = 5. Then , the number is 345.