Find the unit's digit in the product $(2467)^{153} \times (341)^{72}$.

### Concept & Formula To find the unit digit of a number with a large exponent, use the cyclicity rule. Extract the unit digit of the base. - The cyclicity of 7 is 4 (it repeats the pattern 7, 9, 3, 1). - The cyclicity of 1 is 1 (it is always 1 regardless of the power). ### Step-by-Step Solution - **Given:** We need to find the unit digit of $7^{153} \times 1^{72}$. - **Calculation (Base 1):** For $1^{72}$, the unit digit is inherently 1 because 1 multiplied by itself any number of times remains 1. - **Calculation (Base 7):** For $7^{153}$, we divide the exponent 153 by the cyclicity of 4 to find the position in the cycle. - $153 \div 4$ yields a quotient of 38 and a remainder of 1. - A remainder of 1 corresponds to the 1st power in the cycle: $7^1 = 7$. - **Final Product:** Multiply the resulting unit digits from both parts: $7 \times 1 = 7$. ### Exam Strategy & Shortcut To quickly find the remainder when dividing an exponent by 4, you only need to divide the last two digits of the exponent. For 153, just divide 53 by 4. $53 = 13 \times 4 + 1$. The remainder is 1. This instantly tells you to evaluate $7^1$. ### Common Pitfall A common mistake is taking the remainder of the base (dividing 2467 by 4) instead of the exponent. The cyclicity rule dictates that the exponent is divided by the length of the cycle, while the base is stripped down to just its unit digit. ### Final Answer Therefore, the correct answer is **7**.