Question 13

Step 1 · Find the Unit's and Ten's Digits

Form groups of three digits from the right: $\text{Group 2: } 1, \quad \text{Group 1: } 331$

Unit's digit: $\text{Group 1 } (331) \text{ ends in } 1 \implies \text{Unit's digit} = 1$

Ten's digit: $\text{Group 2 is } 1$ $1^3 = 1 \le 1 \implies \text{Ten's digit} = 1$

$\sqrt[3]{1331} = 11$

Step 1 · Find the Unit's and Ten's Digits

Form groups of three digits from the right: $\text{Group 2: } 4, \quad \text{Group 1: } 913$

Unit's digit: $\text{Group 1 } (913) \text{ ends in } 3$ $\text{Since } 7^3 = 343 \text{ ends in } 3 \implies \text{Unit's digit} = 7$

Ten's digit: $\text{Group 2 is } 4$ $1^3 = 1 \quad \text{and} \quad 2^3 = 8$ $\text{Since } 1^3 \le 4 < 2^3 \implies \text{Ten's digit} = 1$

$\sqrt[3]{4913} = 17$

Step 1 · Find the Unit's and Ten's Digits

Form groups of three digits from the right: $\text{Group 2: } 12, \quad \text{Group 1: } 167$

Unit's digit: $\text{Group 1 } (167) \text{ ends in } 7$ $\text{Since } 3^3 = 27 \text{ ends in } 7 \implies \text{Unit's digit} = 3$

Ten's digit: $\text{Group 2 is } 12$ $2^3 = 8 \quad \text{and} \quad 3^3 = 27$ $\text{Since } 2^3 \le 12 < 3^3 \implies \text{Ten's digit} = 2$

$\sqrt[3]{12167} = 23$

Step 1 · Find the Unit's and Ten's Digits

Form groups of three digits from the right: $\text{Group 2: } 32, \quad \text{Group 1: } 768$

Unit's digit: $\text{Group 1 } (768) \text{ ends in } 8$ $\text{Since } 2^3 = 8 \implies \text{Unit's digit} = 2$

Ten's digit: $\text{Group 2 is } 32$ $3^3 = 27 \quad \text{and} \quad 4^3 = 64$ $\text{Since } 3^3 \le 32 < 4^3 \implies \text{Ten's digit} = 3$

$\sqrt[3]{32768} = 32$