Consider the eight digit number 93217648. Using only the second, fourth and sixth digits of this number, and using each of these three digits exactly once, it is possible to form a three digit number that is the square of a two digit odd number. Which of the following is that two digit odd number?

Introduction / Context:
This question combines digit selection with knowledge of perfect squares. You must first identify the specific digits from the given large number, then use them to construct possible three digit numbers. One of these numbers is a perfect square of a two digit odd integer. Finally, you need to identify which two digit odd number produces that square. This type of reasoning problem strengthens your ability to handle constraints and number properties simultaneously.

Given Data / Assumptions:

• The given number is 93217648.
• We are allowed to use only the second, fourth and sixth digits of this number.
• We must use each of these three digits exactly once to form a three digit number.
• The resulting three digit number is the square of a two digit odd number.
• The candidate two digit odd numbers in the options are 17, 19, 15, 21 and 23.

Concept / Approach:
The approach is to first extract the required digits by position, then list all permutations of those digits to see which three digit numbers can be formed. Separately, we find the squares of the given two digit odd numbers and check whether any of these squares match a number that can be formed from the chosen digits. Using properties of squares and systematic checking of permutations ensures that we do not miss the correct answer.

Step-by-Step Solution:
Write the digits of 93217648 with positions: 1: 9, 2: 3, 3: 2, 4: 1, 5: 7, 6: 6, 7: 4, 8: 8. The second digit is 3, the fourth digit is 1, and the sixth digit is 6. So the three digits available are 3, 1 and 6. List all three digit numbers that can be formed using 3, 1 and 6 exactly once: 136, 163, 316, 361, 613 and 631. Now compute the squares of the given two digit odd numbers: 17^2 = 289, 19^2 = 361, 15^2 = 225, 21^2 = 441 and 23^2 = 529. Compare these squares with the list of three digit permutations. Only 361 appears in the list and is the square of 19. Therefore the required two digit odd number whose square can be formed from the digits 3, 1 and 6 is 19.

Verification / Alternative check:
To verify, note that 361 uses digits 3, 6 and 1 exactly once each, which is consistent with the constraint. Also, none of the other squares 289, 225, 441 or 529 can be formed from the digits 3, 1 and 6 because they contain digits 2, 4, 5 or 9 that are not allowed. Another quick check is to observe that 361 is a well known perfect square of 19, which further confirms our result. Hence there is no alternative match and 19 is uniquely correct.

Why Other Options Are Wrong:
Option 17 is wrong because 17^2 = 289, which uses digits 2, 8 and 9 that are not in the allowed set {1, 3, 6}. Option 15 is wrong because 15^2 = 225, which repeats the digit 2 and does not use the digit 3 at all. Option 21 is wrong because 21^2 = 441, which involves the digit 4 twice and again ignores the digits 3 and 6. Option 23 is wrong because 23^2 = 529, containing digits 5, 2 and 9, which are not the required digits.

Common Pitfalls:
A common mistake is to attempt to square random two digit odd numbers other than those supplied in the options, which wastes time. Another pitfall is to misread the positions of digits in the original number or to forget that each of the three digits must be used exactly once. Some candidates also forget to check all permutations of the digits and may wrongly discard 361. Working methodically through positions, permutations and squares is the safest approach.

Final Answer:
The two digit odd number whose square can be formed using the second, fourth and sixth digits of 93217648 is 19.