Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
(i) We have the prime factors of 81 as:
\(\qquad \begin{array}{c|lcr}
3 & 81\\
\hline
3 & 27\\
\hline
3 & 9\\
\hline
3 & 3\\
\hline
& 1
\end{array}
\)
\(81 = 3 \times 3 \times 3 \times 3 = 3^3 \times 3\)
We can observe clearly that, number 3 is divided to 81 to make it a perfect cube.
So, 81 ÷ 3 = 27 which is a perfect cube.
Thus, the required smallest number to be divided is 3.
(ii) We have the prime factors of 256 as:
\(\qquad \begin{array}{c|lcr}
2 & 128\\
\hline
2 & 64\\
\hline
2 & 32\\
\hline
2 & 16\\
\hline
2 & 8\\
\hline
2 & 4\\
\hline
2 & 2\\
\hline
& 1\\
\end{array}
\)
\(128 =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3 \times 2\)
We can observe clearly that, number 2 is divided to 128 to make it a perfect cube.
So, 128 ÷ 2 = 64 which is a perfect cube.
Thus, the required smallest number to be divided is 2.
(iii) We have the prime factors of 135 as:
\(\qquad \begin{array}{c|lcr}
3 & 135\\
\hline
3 & 45\\
\hline
3 & 15\\
\hline
5 & 5\\
\hline
& 1
\end{array}
\)
\(81 = 3 \times 3 \times 3 \times 5 = 3^3 \times 5\)
We can observe clearly that, number 5 is divided to 135 to make it a perfect cube.
So, 135 ÷ 5= 27 which is a perfect cube.
Thus, the required smallest number to be divided is 5.
(iv) We have the prime factors of 192 as:
\(\qquad \begin{array}{c|lcr}
2 & 192\\
\hline
2 & 96\\
\hline
2 & 48\\
\hline
2 & 24\\
\hline
2 & 12\\
\hline
2 & 6\\
\hline
3 & 3\\
\hline
& 1\\
\end{array}
\)
\(192 =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^3 \times 2^3 \times 3\)
We can observe clearly that, number 3 is divided to 192 to make it a perfect cube.
So, 192 ÷ 3 = 64 which is a perfect cube.
Thus, the required smallest number to be divided is 3.
(v) We have the prime factors of 704 as:
\(\qquad \begin{array}{c|lcr}
2 & 704\\
\hline
2 & 352\\
\hline
2 & 176\\
\hline
2 & 88\\
\hline
2 & 44\\
\hline
2 & 22\\
\hline
11 & 11\\
\hline
& 1\\
\end{array}
\)
\(128 =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 = 2^3 \times 2^3 \times 11\)
We can observe clearly that, number 11 is divided to 704 to make it a perfect cube.
So, 704 ÷ 11 = 64 which is a perfect cube.
Thus, the required smallest number to be divided is 11.