Factorials & Power Cycles Questions and Answers

Answer: C

The unit's digit is 6.

The unit's digit of \(47^{25}\) depends only on the unit's digit 7. Powers of 7 end in a repeating cycle of four: 7, 9, 3, 1. Since \(25 = 4 \times 6 + 1\), the 25th power is in position 1 of the cycle and ends in 7.

Subtracting 1 gives a number ending in \(7 - 1 = 6\).

7 is the unit's digit of \(47^{25}\) itself, before the 1 is subtracted, and 1 and 8 come from using the wrong position in the cycle.

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.

Answer: B

\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

24 is \( 4! \) and 720 is \( 6! \), the neighbours on either side.

Answer: D

The unit digits of powers of 7 cycle with period 4:

\[ 7,\ 9,\ 3,\ 1 \]

The fourth power lands on the last of the cycle, so the unit digit is 1. Indeed \( 7^{4} = 2401 \).

Answer: A

For any integer \(n \ge 5\), the factorial \(n!\) will contain the factors 2 and 5. The product of 2 and 5 is 10. Any number multiplied by 10 will have a unit digit of 0. Since 13 is greater than 5, \(13!\) will have a unit digit of 0.

Answer: B

The power cycle of the unit digit of 2 is (2, 4, 8, 6), which has a length of 4. To find the unit digit of \(2^{10}\), we find the remainder of the power when divided by the cycle length: \(10 \div 4\) gives a remainder of 2. The required unit digit is the 2nd in the cycle, which is 4.

Answer: C

The number of trailing zeros in a factorial is determined by the number of times 5 is a factor in its prime factorization. We can find this using the formula: \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + ...\). For 25!, this is \(\lfloor 25/5 \rfloor + \lfloor 25/25 \rfloor = 5 + 1 = 6\). There are 6 trailing zeros.

Answer: B

\[ 1 + 2 + 6 + 24 = 33 \]

The unit digit is 3. Only the unit digits of the terms matter: \( 1 + 2 + 6 + 4 = 13 \), ending in 3.