1 step) a. across should be a prime number, which digit sum (c. across) is a 1 digit number.
2 step) c. across should be an odd number, because b. down is a prime number and can't be divisible by 2.
3 step) from step 2 -> a. across should be a prime number, which digit sum (c. across) is odd number.
4 step) from 1 and 3 steps -> a. across first digit should be even, because second digit is always odd for their sum to be odd.
5 step) from 1 and 4 steps -> we get two numbers which satisfy the rules, it is 43 or 61. So in b. down we get 37 or 17.
6 step) a. down is a square of the sum of the digits of b. down, so from step 5 -> we get a. down is 100 or 64.
7 step) 100 doesn't fit to our rules, 64 does.
Answer:
6 1
4 7
Yes, your attempt to the problem is correct. here is another way of solving it.
The equation of a circle is $$x^2 + y^2 = R^2\Rightarrow x^2 + y^2 = (\frac{d}2)^2$$
where, R and d are the radius and the diameter of the circle respectively.
So, the equation for the top semi circle is $$y = \sqrt{(\frac{d}2)^2 - x^2}$$
Let, the diameter of the circle be of the form $$d = 10a + b$$
where, $a$ and $b$ are whole numbers between 0 and 9 inclusive.
So, $a$ and $b$ are the digits of the two digit number.
Also, it is given that the length of cord CD is obtained by reversing the digits of the diameter. So, $$CD = 10b + a$$
Let us define a function $C(x)$ which gives us the length of a perpendicular chord, whose input is the distance of the center of the chord from the center of the circle.
We know that the length of half of the perpendicular chord is given by $$\sqrt{(\frac{d}2)^2 - x^2}$$
So,$$C(x) = 2\sqrt{(\frac{d}2)^2 - x^2}\Rightarrow 2\sqrt{(\frac{10a + b}2)^2 - x^2}$$
Now, let the distance of CD from the center of the circle be a rational number of the form $\frac{p}q$, where $p$ and $q$ are relatively prime integers.
Hence, the length of CD is given by $$C(\frac{p}q) = 2\sqrt{(\frac{10a + b}2)^2 - (\frac{p}q)^2}$$$$\Rightarrow10b + a = 2\sqrt{(\frac{10a + b}2)^2 - (\frac{p}q)^2}$$
Simplifying the equation will give us,$$(\frac{2p}{3q})^2 = 11(a^2-b^2)$$Since, the LHS is a perfect square, RHS must also be a perfect square. As, the RHS is already a multiple of $11$, $ a^2-b^2$ must be an odd power of 11, like $11^1, 11^2, 11^3$, etc.
Since, $a$ and $b$ are whole numbers between 0 and 9, the maximum value of $a^2-b^2$ will be when $a = 9$ and $b = 0$ , i.e. $81$.
Because, all odd powers of $11$ are greater than $81$ except $11$ itself, $a^2 - b^2$ must be $11$.
So, when we see a list of perfect squares from $1^2$ to $9^2$ $$1, 4, 9, 16, 25, 36, 49, 64, 81$$we see that only $36\;(i.e.\;6^2)$ and $25\;(i.e.\;5^2)$ are $11$ apart. Hence, we can choose, $a = 6$ and $b = 5$, to satisfy the equation $a^2 - b^2 = 11$.
Therefore, the length of the diameter of the circle is $65$ units.
Best Answer
In response to your question: while there are many solutions where the squares have an odd number of digits, there are none for squares with an even number of digits such as 4.
Proof: Let $n \neq r$ be the number and its reverse, and let $10^{2k+1} < n^2 \neq r^2 < 10^{2k+2}$. Then $r^2$ ends with 1, 2, 5, 6 or 9 (not a 0, since then $n$ would start with 0). However, if $r^2$ ends with 1 then $n^2$ must start with 1, meaning $3.16 \times 10^{k} < n < 4.47 \times 10^{k}$; hence $n$ must start with either 3 or 4; but then $r^2$ ends with 9 or 6, a contradiction. By the same logic, if $r^2$ ends with 2 then $n$ starts with 4 or 5, meaning $r^2$ ends with 6 or 5. If $r^2$ ends with 5, $n$ starts with 7. If $r^2$ ends with 6, $n$ starts with 7 or 8. And if $r^2$ ends with 9, $n$ starts with 9. Since none of these are possible, no solution exists.
@tong_nor's answer demonstrates an infinite sequence of examples where the squares have an odd number of digits. In fact, there are many other such examples. A quick Python script suggests that all the the digits of the roots are always in the range 0 to 3. Proving this may be a good follow uo question.