Determine which positive integers a make 10^a divisible by a³+a²+a+1, and find their product | Step-by-Step Solution

What We're Solving:

We need to find all positive integers a where 10^a is divisible by the expression a³+a²+a+1, then multiply all those values of a together.

The Approach:

This is a fantastic number theory problem! We're going to:

• 1. Factor the expression a³+a²+a+1 to better understand its structure
• 2. Analyze what it means for 10^a to be divisible by this expression
• 3. Use properties of divisibility and prime factorization to find constraints on a
• 4. Test small values systematically

The key insight is that 10^a = 2^a × 5^a, so we need a³+a²+a+1 to divide this product.

Step-by-Step Solution:

Step 1: Factor a³+a²+a+1 a³+a²+a+1 = a²(a+1) + 1(a+1) = (a²+1)(a+1)

Step 2: Set up the divisibility condition We need: (a²+1)(a+1) divides 10^a = 2^a × 5^a

This means both factors a²+1 and a+1 must have their prime factors come from {2, 5} only.

Step 3: Analyze a+1 Since a+1 must divide 2^a × 5^a, and a+1 can only have prime factors 2 and 5, we can write: a+1 = 2^i × 5^j for some non-negative integers i, j

Step 4: Analyze a²+1 Similarly, a²+1 must also only have prime factors 2 and 5.

Step 5: Test small values systematically For a = 1:

• a+1 = 2 ✓ (only factor 2)
• a²+1 = 2 ✓ (only factor 2)
• a³+a²+a+1 = 4
• Does 4 divide 10¹ = 10? No, since 10 = 2×5 and we need 2².

For a = 2:

• a+1 = 3 ✗ (has factor 3, which isn't 2 or 5)

For a = 3:

• a+1 = 4 = 2² ✓
• a²+1 = 10 = 2×5 ✓
• a³+a²+a+1 = 40
• Does 40 divide 10³ = 1000? Yes! 1000 ÷ 40 = 25 ✓

For a = 4:

• a+1 = 5 ✓
• a²+1 = 17 ✗ (17 is prime, not 2 or 5)

For larger values of a, a²+1 grows quickly and becomes increasingly likely to have prime factors other than 2 and 5.

Step 6: Verify our solution We found a = 3 works. Let's double-check:

• When a = 3: a³+a²+a+1 = 27+9+3+1 = 40
• 10³ = 1000
• 1000 ÷ 40 = 25 ✓

The Answer:

The only positive integer a that satisfies our condition is a = 3.

Therefore, the product of all such positive integers is simply 3.

Memory Tip:

When dealing with divisibility problems involving expressions like a^n + terms, try factoring first! Also remember that 10^a = 2^a × 5^a, so any divisor can only have prime factors 2 and 5. This severely limits our possibilities and makes systematic checking much more manageable.