1. **Problem statement:** We want to prove that for all integers $n \geq 24$, there exist nonnegative integers $k$ and $l$ such that $$n = 5k + 7l.$$ 2. **Method:** We will use complete (strong) induction on $n$. 3. **Base cases:** Check $n=24, 25, 26, 27, 28$ (since the largest coefficient is 7, we check the first 7 values starting at 24). - $24 = 5 \times 1 + 7 \times 3$ (with $k=1, l=3$) - $25 = 5 \times 5 + 7 \times 0$ (with $k=5, l=0$) - $26 = 5 \times 2 + 7 \times 2$ (with $k=2, l=2$) - $27 = 5 \times 4 + 7 \times 1$ (with $k=4, l=1$) - $28 = 5 \times 0 + 7 \times 4$ (with $k=0, l=4$) All base cases hold. 4. **Inductive hypothesis:** Assume for all integers $m$ with $24 \leq m \leq n$, there exist nonnegative integers $k, l$ such that $$m = 5k + 7l.$$ 5. **Inductive step:** We want to prove the statement for $n+1$. Since $n+1 - 5 \geq 24$ (because $n \geq 24$), by the inductive hypothesis, there exist $k', l'$ such that $$n+1 - 5 = 5k' + 7l'.$$ Adding 5 to both sides gives $$n+1 = 5(k' + 1) + 7l'.$$ Since $k' + 1$ and $l'$ are nonnegative integers, the statement holds for $n+1$. 6. **Conclusion:** By the principle of complete induction, for all integers $n \geq 24$, there exist nonnegative integers $k$ and $l$ such that $$n = 5k + 7l.$$