FACTORS AND FACTORIALS(LEVEL-1)
Q1–10 of 50Find the number of divisors of 1728.
Prime factorization: $1728 = 12^3 = (2^2 \times 3)^3 = 2^6 \times 3^3$
Total divisors = $(6 + 1)(3 + 1) = 7 \times 4 = 28$
Answer: (c) 28
Find the number of divisors of 1080 excluding the divisors, which are perfect squares.
Prime factorization: $1080 = 2^3 \times 3^3 \times 5^1$
Total divisors = $(3 + 1)(3 + 1)(1 + 1) = 32$
Divisors that are perfect squares must have even exponents in their prime factorization ($2^a \cdot 3^b \cdot 5^c$ where $a \in \{0, 2\}$, $b \in \{0, 2\}$, $c \in \{0\}$):
Count = $2 \times 2 \times 1 = 4$
Divisors excluding perfect squares = $32 - 4 = 28$
Answer: (a) 28
Find the number of divisors of 544 excluding 1 and 544.
Prime factorization: $544 = 2^5 \times 17^1$
Total divisors = $(5 + 1)(1 + 1) = 12$
Excluding 1 and the number itself (2 divisors) = $12 - 2 = 10$
Answer: (d) 10
Find the number of divisors of 544 which are greater than 3.
Divisors of 544 are: 1, 2, 4, 8, 16, 17, 32, 34, 68, 136, 272, 544 (12 total).
Divisors $\le 3$ are: 1, 2 (2 divisors).
Divisors $> 3$ = $12 - 2 = 10$
Answer: (b) 10
Find the sum of divisors of 544 excluding 1 and 544.
Formula for sum of divisors $\sigma(n) = \frac{2^{5+1}-1}{2-1} \times \frac{17^{1+1}-1}{17-1} = 63 \times 18 = 1134$
Exclude 1 and 544: $1134 - 1 - 544 = 589$
Answer: (c) 589
Find the sum of divisors of 544 which are perfect squares.
Perfect square divisors of $544 = 2^5 \times 17^1$ are $2^0, 2^2, 2^4$ (which are 1, 4, 16).
Sum = $1 + 4 + 16 = 21$
Answer: (d) 21
Find the sum of odd divisors of 544.
$544 = 2^5 \times 17^1$. Odd divisors come from setting the power of 2 to $2^0$.
Odd divisors: $17^0 = 1$ and $17^1 = 17$.
Sum = $1 + 17 = 18$
Answer: (a) 18
Find the sum of even divisors of 4096.
Prime factorization: $4096 = 2^{12}$
Even divisors: $2^1, 2^2, \dots, 2^{12}$
Sum = $2^1 + 2^2 + \dots + 2^{12} = 2(2^{12} - 1) = 2(4095) = 8190$
Answer: (c) 8190
Find the sum the sums of divisors of 144 and 160.
Sum of divisors of $144 = 2^4 \times 3^2$:
$$\sigma(144) = (1+2+4+8+16)(1+3+9) = 31 \times 13 = 403$$Sum of divisors of $160 = 2^5 \times 5^1$:
$$\sigma(160) = (1+2+4+8+16+32)(1+5) = 63 \times 6 = 378$$Combined sum = $403 + 378 = 781$
Answer: (b) 781
Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600.
Even divisors of $96 = 2^5 \times 3^1$:
Odd divisors of $3600 = 2^4 \times 3^2 \times 5^2$:
Combined sum = $248 + 403 = 651$
Answer: (c) 651