Doubt 1

Solution: 26,244 Divided to Make a Perfect Cube

Question: Divide 26,244 by the smallest number so that the quotient is a perfect cube. Also find the cube root of the quotient.

Step 1: Prime Factorisation of 26,244

We repeatedly divide 26,244 by the smallest prime numbers:

\[ 26244 \div 2 = 13122 \] \[ 13122 \div 2 = 6561 \] \[ 6561 \div 3 = 2187 \] \[ 2187 \div 3 = 729 \] \[ 729 \div 3 = 243 \] \[ 243 \div 3 = 81 \] \[ 81 \div 3 = 27 \] \[ 27 \div 3 = 9 \] \[ 9 \div 3 = 3 \] \[ 3 \div 3 = 1 \]

Therefore,

\[ 26244 = 2^2 \times 3^8 \]

Step 2: Condition for a Perfect Cube

In the prime factorisation of a perfect cube, the powers of all prime factors must be multiples of 3.

For example:

\[ 8 = 2^3 \] \[ 27 = 3^3 \]

But in

\[ 26244 = 2^2 \times 3^8 \]

the powers are \(2\) and \(8\), neither of which is a multiple of \(3\).

Step 3: Find the Smallest Divisor

We have to divide the number so that the remaining powers become multiples of \(3\).

For \(2^2\), we remove \(2^2\):

\[ 2^2 \div 2^2 = 2^0 \]

For \(3^8\), we remove \(3^2\):

\[ 3^8 \div 3^2 = 3^6 \]

Since \(6\) is a multiple of \(3\), the remaining number will be a perfect cube.

Therefore, the smallest number is:

\[ 2^2 \times 3^2 \] \[ = 4 \times 9 \] \[ = \mathbf{36} \]

Step 4: Find the Quotient

\[ 26244 \div 36 = 729 \]

Alternatively, using prime factors:

\[ \frac{2^2 \times 3^8}{2^2 \times 3^2} \] \[ = 3^{8-2} \] \[ = 3^6 \] \[ = (3^2)^3 \] \[ = 9^3 \] Therefore,

\[ \boxed{\text{Quotient} = 729} \]

Step 5: Find the Cube Root

We need to find:

\[ \sqrt[3]{729} \]

Since

\[ 9 \times 9 \times 9 = 729 \] therefore, \[ \sqrt[3]{729} = \mathbf{9} \]

Final Answer

Verification

\[ 36 \times 729 = 26244 \] and \[ 9^3 = 729 \]