Maths 2 Mark Question

On factorising 456533 into prime factors, we get
456533 = 7 × 7 × 7 × 11 × 11 × 11
Group the factors in triples of equal factors as:
456533 = {7 × 7 × 7} × {11 × 11 × 11}
It is evident that the prime factors of 456533 can be grouped into triples of equal factors and no factor is left over.
Therefore, 456533 is a perfect cube.

On factorising 7803 into prime factors, we get:
7803 = 3 × 3 × 3 × 17 × 17
On grouping the factors in triples of equal factors, we get:
7803 = {3 × 3 × 3} × 17 × 17
It is evident that the prime factors of 7803 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 7803 is a not perfect cube. However, if the number is multiplied by 17, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 7803 should be multiplied by 17 to make it a perfect cube.

On factorising 1600 into prime factors, we get:
1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
On grouping the factors in triples of equal factors, we get:
1600 = {2 × 2 × 2} × {2 × 2 × 2} × 5 × 5
It is evident that the prime factors of 1600 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 1600 is a not perfect cube. However, if the number is divided by (5 × 5 = 255 × 5 = 25), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 1600 should be divided by 25 to make it a perfect cube.

On factorising 8640 into prime factors, we get:
8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
On grouping the factors in triples of equal factors, we get:
8640 = {2 × 2 × 2} × {2 × 2 × 2} × {3 × 3 × 3} × 5
It is evident that the prime factors of 8640 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8640 is a not perfect cube. However, if the number is divided by 5, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 8640 should be divided by 5 to make it a perfect cube.

On factorising 243000 into prime factors, we get:
243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
On grouping the factors in triples of equal factors, we get:
243000 = {2 × 2 × 2} × {3 × 3 × 3} × 3 × 3 × {5 × 5 × 5}
It is evident that the prime factors of 243000 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243000 is a not perfect cube. However, if the number is divided by (3 × 3 = 9), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 243000 should be divided by 9 to make it a perfect cube.

On factorising 1728 into prime factors, we get:
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
On grouping the factors in triples of equal factors, we get:
1728 = {2 × 2 × 2} × {2 × 2 × 2} × {3 × 3 × 3}
It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.
Thus, (iii) 243 is the required number, which is not a perfect cube.

On factorising 675 into prime factors, we get:
1323 = 3 × 3 × 3 × 7 × 7
On grouping the factors in triples of equal factors, we get:
1323 = {3 × 3 × 3} × 5 × 5
It is evident that the prime factors of 1323 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 1323 is a not perfect cube. However, if the number is multiplied by 7, the factors can be grouped into triples of equal factors and no factor will be left over.
Thus, 1323 should be multiplied by 7 to make it a perfect cube.

On factorising 243 into prime factors, we get
243 = 3 × 3 × 3 × 3 × 3
Group the factors in triples of equal factors as:
243 = {3 × 3 × 3} × 3 × 3
It is evident that the prime factors of 243 can be grouped into triples of equal factors and no factor is left over.
Therefore, 243 is a perfect cube.

On factorising 8788 into prime factors, we get:
8788 = 2 × 2 × 13 × 13 × 13
On grouping the factors in triples of equal factors, we get:
8788 = 2 × 2 × {13 × 13 × 13}
It is evident that the prime factors of 8788 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8788 is a not perfect cube. However, if the number is divided by (2 × 2 = 42 × 2 = 4), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 8788 should be divided by 4 to make it a perfect cube.

On factorising 675 into prime factors, we get:
675 = 3 × 3 × 3 × 5 × 5
On grouping the factors in triples of equal factors, we get:
675 = {3 × 3 × 3} × 5 × 5
It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 675 is a not perfect cube. However, if the number is multiplied by 5, the factors can be grouped into triples of equal factors and no factor will be left over.
Thus, 675 should be multiplied by 5 to make it a perfect cube.

On factorising 3087 into prime factors, we get
3087 = 3 × 3 × 7 × 7 × 7
Group the factors in triples of equal factors as:
3087 = 3 × 3 × {7 × 7 × 7}
It is evident that the prime factors of 3087 can be grouped into triples of equal factors and no factor is left over.
Therefore, 243 is a perfect cube.

On factorising 2560 into prime factors, we get:
2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5
On grouping the factors in triples of equal factors, we get:
2560 = {2 × 2 × 2} × {2 × 2 × 2} × {2 × 2 × 2} × 5
It is evident that the prime factors of 2560 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 2560 is a not perfect cube. However, if the number is multiplied by 5 × 5 = 255 × 5 = 25, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 2560 should be multiplied by 25 to make it a perfect cube.

On factorising 675 into prime factors, we get:
675 = 3 × 3 × 3 × 5 × 5
On grouping the factors in triples of equal factors, we get:
675 = {3 × 3 × 3} × 5 × 5
It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 675 is a not perfect cube. However, if the number is divided by 5 × 5 = 255 × 5 = 25, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 675 should be divided by 25 to make it a perfect cube.

On factorising 35721 into prime factors, we get:
35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
On grouping the factors in triples of equal factors, we get:
35721 = {3 × 3 × 3} × {3 × 3 × 3} × 7 × 7
It is evident that the prime factors of 35721 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 35721 is a not perfect cube. However, if the number is divided by (7 × 7 = 497), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 35721 should be divided by 49 to make it a perfect cube.

On factorising 1728 into prime factors, we get
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
Group the factors in triples of equal factors as:
1728 = {2 × 2 × 2} × {2 × 2 × 2} × {3 × 3 × 3}
It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over.
Therefore, 1728 is a perfect cube.

On factorising 216 into prime factors, we get
216 = 2 × 2 × 2 × 3 × 3 × 3
Group the factors in triples of equal factors as:
216 = {2 × 2 × 2} × {3 × 3 × 3}
It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over.
Therefore, 216 is a perfect cube.

On factorising 243 into prime factors, we get:
243 = 3 × 3 × 3 × 3 × 3
On grouping the factors in triples of equal factors, we get:
243 = {3 × 3 × 3} × 3 × 3
It is evident that the prime factors of 243 can be grouped into triples of equal factors and no factor is left over. Therefore, 243 is a perfect cube.

On factorising 35721 into prime factors, we get:
35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
On grouping the factors in triples of equal factors, we get:
35721 = {3 × 3 × 3} × {3 × 3 × 3} × 7 × 7
It is evident that the prime factors of 35721 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 35721 is a not perfect cube. However, if the number is multiplied by 7, the factors be grouped into triples of equal factors such that no factor is left over.
Thus, 35721 should be multiplied by 7 to make it a perfect cube.

On factorising 64 into prime factors, we get:
64 = 2 × 2 × 2 × 2 × 2 × 2
On grouping the factors in triples of equal factors, we get:
64 = {2 × 2 × 2} × {2 × 2 × 2}
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.

On factorising 166375 into prime factors, we get
166375 = 5 × 5 × 5 × 11 × 11 × 11
Group the factors in triples of equal factors as:
166375 = {5 × 5 × 5} × {11 × 11 × 11}
It is evident that the prime factors of 166375 can be grouped into triples of equal factors and no factor is left over.
Therefore, 166375 is a perfect cube.

On factorising 64 into prime factors, we get
64 = 2 × 2 × 2 × 2 × 2 × 2
Group the factors in triples of equal factors as:
64 = {2 × 2 × 2} × {2 × 2 × 2}
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over.
Therefore, 64 is a perfect cube.

On factorising 107811 into prime factors, we get:
107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
On grouping the factors in triples of equal factors, we get:
107811 = {3 × 3 × 3} × 3 × {11 × 11 × 11}
It is evident that the prime factors of 107811 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 107811 is a not perfect cube. However, if the number is divided by 3, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 107811 should be divided by 3 to make it a perfect cube.

On factorising 107811 into prime factors, we get:
107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
On grouping the factors in triples of equal factors, we get:
107811 = {3 × 3 × 3} × 3 × {11 × 11 × 11}
It is evident that the prime factors of 107811 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 107811 is a not perfect cube. However, if the number is multiplied by 3 × 3 = 93 × 3 = 9, the factors be grouped into triples of equal factors such that no factor is left over.
Thus, 107811 should be multiplied by 9 to make it a perfect cube.