Express 9801 as a product of its prime factors. Answer --------------------------------------------------------- Hence, explain why 9801 is a square number. Answer --------------------------------------------------------- a and b are both prime numbers. Find the smallest value of a and b such that $9801\times\frac{a}{b}$ is a perfect cube. Answer a = ___ b = ___

1. Divide repeatedly by prime factors until only primes remain.

$9801\div 3=3267$

$3267\div 3=1089$

$1089\div 3=363$

$363\div 3=121$

$121=11\times 11$

2. Write the prime factors in index form.

Hence,

$9801=3\times 3\times 3\times 3\times 11\times 11=3^4\times 11^2.$

1. Use the prime factorisation to show that 9801 can be written as a square.

From part (i),
$9801=3^4\times 11^2=(3^2\times 11)^2=99^2.$
Hence, $9801$ is a square number.

1. Use the prime factorisation of 9801 and compare indices with the condition for a perfect cube.

From part (i),
$9801=3^4\times 11^2.$
For a perfect cube, every prime index must be a multiple of $3$.

2. Choose prime values of a and b that adjust the indices to multiples of 3.

$9801\times\frac{a}{b}=3^4\times 11^2\times\frac{a}{b}.$
To make the index of $3$ a multiple of $3$, multiply by one more $3$:
$a=3.$
To make the index of $11$ a multiple of $3$, divide by one $11$:
$b=11.$
Then
$9801\times\frac{3}{11}=3^5\times 11,$
This is not a perfect cube, so instead check by adjusting to multiples correctly:
$3^4\times 11^2\times\frac{11}{3}=3^3\times 11^3,$
which is a perfect cube.
Hence
$a=11,\quad b=3.$