Unit Digit — PYPs

The Setup: We are given a positive integer $p$ such that the unit digit of $p^3$ is exactly $4$. We need to determine the possible unit digits of the expression $(p+3)^3$.

Step 1: Determine the unit digit of $p$.
We evaluate the cubes of all base digits from $0$ to $9$ to see which produces a unit digit of $4$:
$0^3 = 0, 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216, 7^3 = 343, 8^3 = 512, 9^3 = 729$.
The only digit whose cube ends in $4$ is $4$ itself.
Therefore, the unit digit of $p$ must be $4$.

Step 2: Calculate the unit digit of $(p+3)^3$.
Since $p \equiv 4 \pmod{10}$, we substitute this into the target expression:

$$p + 3 \equiv 4 + 3 \equiv 7 \pmod{10}$$

Now, cube this resulting unit digit:

$$7^3 = 343$$

The unit digit of $343$ is $3$. Thus, the only possible unit digit for $(p+3)^3$ is $3$.

Final Answer: 3

The Setup: Unit digit mechanics operate on a strict modulo-10 cyclicity. We don't care about the massive numbers; we only care about the final digit of the base and the remainder of the exponent when divided by its cycle length.
Step 1: Analyze the first term: $743^{85}$.
The unit digit depends entirely on $3^{85}$. The powers of $3$ cycle in a 4-step loop ($3$, $9$, $7$, $1$). Divide the exponent $85$ by $4$ to find the remainder.

$$85=4\times21+1$$

Since the remainder is $1$, it hits the first step of the cycle. The unit digit is $3^1=3$.
Step 2: Analyze the second term: $525^{37}$.
This is a freebie. Any positive integer power of a number ending in $5$ will always end in $5$. The unit digit is $5$.
Step 3: Analyze the third term: $987^{96}$.
The unit digit depends on $7^{96}$. The powers of $7$ also cycle in a 4-step loop ($7$, $9$, $3$, $1$). Divide $96$ by $4$.

$$96=4\times24+0$$

Since it divides perfectly (remainder $0$), it lands on the final step of the cycle. The unit digit is $1$.
Step 4: Combine the unit digits.

$$3-5+1=-1$$

Since we can't output a negative unit digit in a standard integer, we simulate borrowing a $10$ from the next placeholder column over (just like elementary school subtraction).

$$10-1=9$$

Final Answer: 9