Divisibility Rules

Concept: The least number divisible by a set of given numbers is their Least Common Multiple (LCM). Step 1: List the numbers The given numbers are 2, 3, 7, and 9. Step 2: Find the prime factorization of each number



Step 3: Calculate the LCM To find the LCM, take the highest power of each prime factor present in any of the numbers. The prime factors involved are 2, 3, and 7.
Highest power of 2:
Highest power of 3: (from the number 9, as )
Highest power of 7: LCM = LCM = LCM = Calculate : . So, the LCM is 126. The least number divisible by 2, 3, 7, and 9 is 126. Note: Since 9 is a multiple of 3, any number divisible by 9 is automatically divisible by 3. So, we effectively need the LCM of 2, 7, and 9. LCM(2, 7, 9) = LCM(2, 7, ). Since 2, 7, and 9 are pairwise coprime (no common factors other than 1), their LCM is their product: .

Concept: This problem involves testing divisibility of an algebraic expression. We can use algebraic identities or test small values of . The relevant algebraic identity is the difference of squares: . More generally, is divisible by for any positive integer , and divisible by if is an even positive integer. Step 1: Rewrite the expression The given expression is . We can rewrite as . We can rewrite as . So the expression becomes . Using the property that is always divisible by : Here, and . So, is always divisible by . . The divisors of 65 are 1, 5, 13, 65. Therefore, is always divisible by 5 and by 13. Alternative Method using Difference of Squares repeatedly (if n allows) Let and . Then the expression is . So, . Step 2: Test with small values of (since is a natural number, ) Case 1: Let . . Is 65 divisible by 5? Yes, . Is 65 divisible by 13? Yes, . So for , the expression is divisible by both 5 and 13. Case 2: Let . . Using the difference of squares formula : . Since 65 is a factor, and 65 is divisible by both 5 and 13, the expression is also divisible by both 5 and 13. Step 3: General Proof As shown in Step 1, the expression can be written as . We know that for any positive integer , is always divisible by . Here and . So . Since 65 is divisible by 5 (as ) and divisible by 13 (as ), it follows that is always divisible by both 5 and 13.