What is the rightmost digit preceding the zeros in the value of 30³⁰?

The correct option is 9.

To find the rightmost digit preceding the trailing zeros in the value of:
${30}^{30}$
we can rewrite the base using its prime factors:

${30}^{30}={\left(3\times 10\right)}^{30}=3^{30}\times {10}^{30}$

The term ${10}^{30}$ represents a 1 followed by 30 zeros. Multiplying any integer by ${10}^{30}$ appends exactly 30 trailing zeros to that integer. Therefore, the rightmost digit preceding these trailing zeros is determined entirely by the units digit (the last digit) of:
$3^{30}$

Let us examine the units digits of successive powers of 3 to find a pattern:
• $3^1=3$ (units digit is 3)
• $3^2=9$ (units digit is 9)
• $3^3=27$ (units digit is 7)
• $3^4=81$ (units digit is 1)
• $3^5=243$ (units digit is 3)

The units digits repeat in a periodic cycle of length 4: 3, 9, 7, 1.

To find the position in the cycle for the exponent 30, we divide 30 by the cycle length of 4:
$30=\left(4\times 7\right)+2$
This division gives a quotient of 7 and a remainder of 2.

The remainder of 2 indicates that the units digit of $3^{30}$ is the second digit in the repeating cycle:
• First position: 3
• Second position: 9

Thus, the units digit of $3^{30}$ is 9, which means the rightmost digit preceding the zeros in ${30}^{30}$ is 9.