a number which sum of digit is 1977. Prove that the number will not be a perfect square
Hello, Swarnav!
A number whose sum of digits is 1977.
Prove that the number will not be a perfect square.
Note: in this problem, a "number" means a positive integer.
Consider the digital root of a number.
Add the digits together.
If the sum has more than one digit, add its digits together.
Continue until we have a one-digit sum.
That is the digital root of the number.
A number can have only nine possible digital roots.
Examine the digital roots of their squares.
. . $\displaystyle \begin{array}{ccc}\text{digit root} && \text{digital root} \\ \text{of }n & n^2 & \text{of }n^2 \\ \hline 1 & 1 & 1 \\ 2 & 4 & 4 \\ 3 & 9 & 9 \\ 4 & 16 & 7 \\ 5 & 25 & 7 \\ 6 & 36 & 9 \\ 7 & 49 & 4 \\ 8 & 64 & 1 \\ 9 & 81 & 9 \end{array}$
The digital root of a square must be: $\displaystyle 1, 4, 7,\text{ or }9.$
Since the digital root of $\displaystyle 1977$ is $\displaystyle 6$, it cannot be a square.
Hello, Swarnav!
What is your point?But 6660 is not a perfect square, but it digit sum is 6 + 6 + 6 + 0 = 18, then 1 + 8 = 9.
Read my post again . . .
I said: If a number is a square, then its digital root is 1, 4, 7 or 9.
You are taking the converse:
. . "If a number's digital root is 1, 4, 7 or 9, then it must be a square."
I didn't say that.
We cannot trust the converse.
Consider the true statement: "If the figure is a square, then it has four side."
The converse is: "If the figure has four sides, then it is a square."
This statement may or may not be true . . . agreed?