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(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

Answer :

(i) We have the prime factors of 243 as:

\(\qquad \begin{array}{c|lcr}
3 & 243\\
\hline
3 & 81\\
\hline
3 & 27\\
\hline
3 & 9\\
\hline
3 & 3\\
\hline
& 1
\end{array}
\)

\(243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^3 \times 3 \times 3\)

We can observe clearly that, number 3 is required to make \(3 \times 3\) a group of three.
Thus, the required smallest number to be multiplied is 3.

(ii) We have the prime factors of 256 as:

\(\qquad \begin{array}{c|lcr}
2 & 256\\
\hline
2 & 128\\
\hline
2 & 64\\
\hline
2 & 32\\
\hline
2 & 16\\
\hline
2 & 8\\
\hline
2 & 4\\
\hline
2 & 2\\
\hline
& 1\\
\end{array}
\)

\(256 =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3 \times 2 \times 2\)

We can observe clearly that, number 2 is required to make \(2 \times 2 \) a group of three.Thus, the required smallest number to be multiplied is 2.

(iii) We have the prime factors of 72 as:

\(\qquad \begin{array}{c|lcr}
2& 72\\
\hline
2 & 36\\
\hline
2 & 18\\
\hline
3 & 9\\
\hline
3 & 3\\
\hline
& 1
\end{array}
\)

\(72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3 \times 3\)

We can observe clearly that, number 3 is required to make \(3 \times 3\) a group of three.Thus, the required smallest number to be multiplied is 3.

(iv) We have the prime factors of 675 as:

\(\qquad \begin{array}{c|lcr}
3 & 675\\
\hline
3 & 225\\
\hline
3 & 75\\
\hline
5 & 25\\
\hline
5 & 5\\
\hline
& 1
\end{array}
\)

\(675 = 3 \times 3 \times 3 \times 5 \times 5 = 3^3 \times 5 \times 5\)

We can observe clearly that, number 5 is required to make \(5 \times 5\) a group of three.Thus, the required smallest number to be multiplied is 5.

(v) We have the prime factors of 100 as:

\(\qquad \begin{array}{c|lcr}
2 & 100\\
\hline
2 & 50\\
\hline
5 & 25\\
\hline
5 & 5\\
\hline
& 1
\end{array}
\)

\(243 = 2 \times 2 \times 5 \times 5 \)

We can observe clearly that, number 2 and 5 are required to make \(2 \times 2\) and \(5 \times 5\) a group of three.Thus, the required smallest number to be multiplied is \(5 \times 2=10\).

- 1.Which of the following numbers are not perfect cubes? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656
- 3.Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. (i) 81 (ii) 128 (iii) 135 (iv) 92 (v) 704
- 4.Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will be needed to form a cube?

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