A student can arrange her books into groups of 5, groups of 9, or groups of 13, with no books left over in each case. What is the smallest possible total number of books she can have?

Introduction:
This problem tests your understanding of the least common multiple (LCM) in a practical context. The student can divide her books into equal groups of various sizes without any remainder, which implies that the total number of books is a common multiple of these group sizes.

Given Data / Assumptions:

• The student can arrange the books in groups of 5 with none left over.
• She can also arrange them in groups of 9 with none left over.
• She can also arrange them in groups of 13 with none left over.
• We need the smallest possible total number of books.

Concept / Approach:
If a number can be divided exactly by 5, 9, and 13, it must be a common multiple of 5, 9, and 13. The smallest such positive integer is the least common multiple (LCM). By calculating the LCM of 5, 9, and 13, we obtain the smallest possible total number of books.

Step-by-Step Solution:
Prime factorise the numbers: 5 is prime. 9 = 3^2 13 is prime. The LCM is the product of all primes with their highest powers present: LCM = 5 × 3^2 × 13 Compute step by step: 3^2 = 9 5 × 9 = 45 45 × 13 = 585 Therefore, the smallest possible number of books is 585

Verification / Alternative check:
Check divisibility: 585 ÷ 5 = 117, 585 ÷ 9 = 65, and 585 ÷ 13 = 45. All results are integers, so the student can indeed arrange her books into groups of 5, 9, or 13 without leftovers. No smaller number can satisfy all three divisibility conditions simultaneously because 585 is the LCM.

Why Other Options Are Wrong:
487, 635, and 705 are not divisible by all three numbers 5, 9, and 13. For instance, 487 is prime and not divisible by 5. 635 is divisible by 5 but not by 9. 705 is divisible by 5 and 9 but not by 13. Hence they cannot be the smallest such total.

Common Pitfalls:
Learners sometimes only check one or two of the divisibility conditions or rely on random trials. Another mistake is to compute the HCF instead of the LCM. Always remember that when a number must be divisible by several given numbers, the LCM is the fundamental tool to use.

Final Answer:
The smallest possible number of books the student can have is 585.