Factorials & Power Cycles Questions and Answers

Answer: B

The product ends in 14 zeroes.

Each trailing zero needs one factor 10 = 2 × 5, so count the 2s and 5s and take the smaller count.

Factors of 5: 5, 5, 10, 15, 15, 20, 35, 40, 45, 55, 80, 85 and 95 give one each (13), and 75, 100 and 100 give two each (6), so there are 19 fives. Factors of 2: 10 gives 1, 20 gives 2, 40 gives 3, 80 gives 4, and each 100 gives 2, so there are \(1 + 2 + 3 + 4 + 2 + 2 = 14\) twos.

The 2s run out first, so only 14 tens can be formed. 18 or 19 comes from counting the 5s, which is the habit from factorial questions where 2s are plentiful; here they are not.

Answer: C

The remainder is 3.

\(8 \equiv 3 \pmod 5\), and the powers of 3 modulo 5 repeat in a cycle of four: 3, 4, 2, 1.

\[ \begin{aligned} 145 &= 4 \times 36 + 1 \\ 8^{145} &\equiv 3^{1} \equiv 3 \pmod 5 \end{aligned} \]

Equivalently, the units digits of powers of 8 cycle 8, 4, 2, 6, and \(8^{145}\) ends in 8, which leaves 3 on division by 5. 6 cannot be a remainder on division by 5, 0 would need a factor 5, and 4 is the next term of the cycle, an off-by-one.

Answer: A

The remainder is 5.

The remainder on division by 10 is just the last digit. Every positive power of 5 ends in 5:

\[ 5^{1} = 5,\ 5^{2} = 25,\ 5^{3} = 125,\ \ldots \]

The tower of powers is some (huge) positive power of 5, so it too ends in 5, and dividing by 10 leaves 5. The exponent's size does not matter, which is why 'Cannot be determined' is wrong. 0 would need the number to end in 0, which needs a factor 2, and powers of 5 have none; 1 is the last-digit cycle of numbers ending in 1.