How many two digit numbers have digits whose sum is a perfect square?
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Two digit integers can be expressed as $a*10 + b*1$ for some integers $a, b$ satisfying $1 \leq a, b \leq 9$. For example, $95$ is $9*10 + 5*1$. Show Now, the "reverse order" integer of $a*10 + b*1$ is $b*10 + a*1$. For example, $59$ is $5*10 + 9*1$, and $95$ is $9*10 + 5*1$. So, from the question, we want $10a + b + 10b + a = c^{2}$ for some $c$. So, we are looking for integers $a$ and $b$ each between $1$ and $9$ so that $10(a + b) + a + b = c^{2}$ for some integer $c$, that is, $11(a + b) = c^{2}$, i.e., $a + b = c^{2}/11$. Maybe from here you can plug in all possible combinations? There would only be 81 to check. Like $a = 1, b = 1$, $a = 1, b = 2$, etc.
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Square Number.. What is the sum of digits that a perfect square?Lemma 3: the repeated sum of the digits of a perfect square is 1,4,7 or 9.
How many 2 digit square numbers are there?There are a total of 6 'two digit square numbers', which can be listed as, 16, 25, 36, 49, 64, and 81.
What is the largest 2 digit number which is a perfect square?Hence largest two digit number who is also a perfect square = 81.
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