Units Digit Patterns in Powers of 4

 

❓ Question

Consider the number $4^n$, where $n$ is a natural number.
Check whether there is any value of $n$ for which $4^n$ ends with the digit 0.

---

🖼️ Solution Image

---

✍️ Short Solution

This is a number system + divisibility logic question.
Focus on prime factors of 10 💯

---

🔹 Step 1 — Condition for Last Digit 0 (MOST IMPORTANT)

If a number ends with 0, then:

$$\text{Number must be divisible by 10}$$
$$10 = 2 \times 5$$

👉 So the number must contain both 2 and 5 as factors

---

🔹 Step 2 — Prime Factorization of $4^n$

$$4 = 2^2$$
$$4^n = (2^2)^n = 2^{2n}$$

👉 Prime factors = only 2

---

🔹 Step 3 — Check Requirement

For last digit 0:

• Factor 2 → Present ✅
• Factor 5 → Not present ❌

👉 Condition not satisfied

---

🔹 Step 4 — Final Conclusion

Since $4^n$ does not contain factor 5,
it can never be divisible by 10

👉 So it will never end with digit 0

---

✅ Final Answer

$$\boxed{\text{There is no natural number } n \text{ for which } 4^n \text{ ends with digit 0}}$$

---

⭐ Key Insight

• Ending with 0 ⇒ divisible by 10
• 10 ⇒ $2 \times 5$
  
• $4^n = 2^{2n}$ ⇒ no factor 5

🧠 Memory Line:

Jisme 5 nahi hoga, wo kabhi 0 pe end nahi karega