Question 11

To make a number a perfect cube, we find its prime factors and ensure each factor appears in groups of three.

Step 1 — Find Prime Factors

We first find the prime factors of the given number, 1323. We divide 1323 by the smallest prime numbers. 1323 is not divisible by 2 because it is an odd number. The sum of its digits ($1+3+2+3 = 9$) is divisible by 3. So, 1323 is divisible by 3. $1323 \div 3 = 441$ The sum of digits of 441 ($4+4+1 = 9$) is divisible by 3. So, 441 is divisible by 3. $441 \div 3 = 147$ The sum of digits of 147 ($1+4+7 = 12$) is divisible by 3. So, 147 is divisible by 3. $147 \div 3 = 49$ 49 is not divisible by 3 or 5. 49 is divisible by 7. $49 \div 7 = 7$ 7 is a prime number. So, the prime factorization of 1323 is: $1323 = 3 \times 3 \times 3 \times 7 \times 7$

$\boxed{1323 = 3^3 \times 7^2}$

Step 2 — Group Factors into Triplets

A perfect cube number has each prime factor appearing three times. We look at the prime factors of 1323. We have three factors of 3, which is $3 \times 3 \times 3$. This forms a complete triplet for the prime factor 3. We have two factors of 7, which is $7 \times 7$. This is not a complete triplet for the prime factor 7.

Step 3 — Calculate the Multiplier

To make 1323 a perfect cube, we need to complete the triplet for 7. The factor 7 appears two times. We need one more 7 to make it $7 \times 7 \times 7$. So, we must multiply 1323 by 7.

$\boxed{\text{Multiplier} = 7}$

Step 4 — Verify the New Cube Number

Let us multiply 1323 by the number we found, which is 7. The new number will be a perfect cube. $1323 \times 7$ $= 9261$ Now, we can check its prime factorization. The new number is $3^3 \times 7^2 \times 7$. This means the new number is $3^3 \times 7^3$. We can write this as $(3 \times 7)^3$. $3 \times 7 = 21$ So, the new number is $21^3$. $21^3 = 21 \times 21 \times 21$ $= 441 \times 21$ $= 9261$ This confirms that 9261 is a perfect cube.

Answer

The number to multiply by 1323 to make it a cube number is 7.