Doubt

Answer: 6

Step 1: Rewrite the expression

The goal is to write $x^2 + 2x + 33$ as

$x^2 + 2x + 33 = (x+3)\cdot(\text{something}) + (\text{a number})$

because then it's easy to see when $x+3$ divides it.

We do this with polynomial long division, one term at a time.

First term. What times $x$ gives $x^2$? Answer: $x$.

$x(x+3) = x^2 + 3x$

Subtract this from the original:

$(x^2 + 2x + 33) - (x^2 + 3x) = -x + 33$

Second term. What times $x$ gives $-x$? Answer: $-1$.

$-1(x+3) = -x - 3$

Subtract again:

$(-x + 33) - (-x - 3) = 36$

Nothing with $x$ is left, so $36$ is the remainder. Putting it together:

$x^2 + 2x + 33 = (x+3)(x-1) + 36$

Quick check: $(x+3)(x-1) = x^2 + 2x - 3$, and $x^2 + 2x - 3 + 36 = x^2 + 2x + 33$. ✓

Shortcut (Remainder Theorem): The remainder when dividing by $x+3$ equals the value at $x = -3$:

$(-3)^2 + 2(-3) + 33 = 9 - 6 + 33 = 36$

Step 2: Use the rewritten form

$(x+3)(x-1)$ is always divisible by $x+3$. So the whole expression is divisible by $x+3$ only if the leftover $36$ is too:

$(x+3) \mid 36$

Step 3: List the divisors

Divisors of $36$: $1, 2, 3, 4, 6, 9, 12, 18, 36$

Since $x \geq 1$, we need $x + 3 \geq 4$, which leaves:

$x + 3 \in {4, 6, 9, 12, 18, 36}$

Step 4: Solve for $x$

$x \in {1, 3, 6, 9, 15, 33}$

So there are $\boxed{6}$ natural numbers.