Question

Question #33200.The sum of the squares of two digits of a positive integral number is 65 and the number is 9 times the sum of its digits. Find the number.
In the solution of Question #33200 you said the possible range is 1 to 8.. I don't understand this possible range.. And in other solutions there is written the number is 10x+y. How does the number is 10x+y.? Please solve this query for me. I'll be waiting for your kind response

Answer 1

Answer on Question #73311 – Math - Algebra

Question

Question #33200. The sum of the squares of two digits of a positive integral number is 65 and the number is 9 times the sum of its digits. Find the number. In the solution of Question #33200 you said the possible range is 1 to 8.. I don't understand this possible range.. And in other solutions there is written the number is $10x + y$. How does the number is $10x + y$? Please solve this query for me. I'll be waiting for your kind response

Solution

Let the first digit be $x$ and the second digit be $y$.

Therefore, the number would be $10x + y$ as the $x$ is the tens digit and $y$ is the ones digit. For example, if the first digit is 5 and the second digit is 9, the number would be 59 which is calculated as

$$5 \cdot 10 + 9 = 59.$$

Sum of the squares of the digits $= x^2 + y^2$

$$65 = x^2 + y^2 \quad \text{(1)}$$

The number is 9 times the sum of its digits:

$$9(x + y) = 10x + y$$ $$9x + 9y = 10x + y$$ $$x = 8y$$

Plugging $x = 8y$ in the equation (1), we get:

$$65 = (8y)^2 + y^2$$ $$65 = 65y^2$$ $$y = 1$$

And $x$ would be $x = 8y$

$$x = 8 \cdot 1 = 8$$

Therefore, the number would be $8 \cdot 10 + 1 = 81$.

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