Quantitative Aptitude - MCQ Practice Questions

A number has exactly 3 factors only when it is the square of a prime.

For p² where p is prime, factors are: 1, p, p².

Example: 4 has factors 1,2,4 (3 factors). 9 has factors 1,3,9 (3 factors).

Divisors of 28: 1, 2, 4, 7, 14, 28.

Sum = 1+2+4+7+14+28 = 56. (Note: 28 is a perfect number where sum of proper divisors = 28)

When a number is expressed in prime factorization form, we use the divisor formula: if $n=p_1^{a_1}\times p_2^{a_2}\times p_3^{a_3}$, then the total number of divisors is $(a_1+1)(a_2+1)(a_3+1)$.

Step 1: Identify the prime factorization

The given number is:

Here, the exponents are: $a_1=3$, $a_2=2$, $a_3=1$

Step 2: Apply the divisor formula

The number of divisors is found by adding 1 to each exponent and multiplying:

Step 3: Calculate

Step 4: Verify with a divisor example

Each divisor has the form $2^a\times 3^b\times 5^c$ where $0\le a\le 3$, $0\le b\le 2$, $0\le c\le 1$. This gives 4 choices for the power of 2, 3 choices for the power of 3, and 2 choices for the power of 5.

**Answer: The total number of divisors is $24$ (Option B)**

Let tens digit = x, units digit = y.

Then x + y = 12 and 10x + y = 6y + 6.

From second: 10x = 5y + 6.

Substituting y = 12-x: 10x = 5(12-x) + 6 = 60 - 5x + 6.

So 15x = 66, x = 8, y = 4.

Number = 84

Using the formula: Product of two numbers = HCF × LCM.

So 180 = 6 × LCM.

Therefore LCM = $\frac{180}{6}$ = 30

We need to find numbers that satisfy two remainder conditions simultaneously. This is a Chinese Remainder Theorem problem.

Step 1: Set up the congruence equations

Given conditions:

From the first condition, we can write:

Step 2: Substitute into the second condition

Substitute $n=9k+4$ into the second congruence:

Step 3: Solve for k

Find the multiplicative inverse of 9 modulo 11. Since $9\times 5=45=44+1\equiv 1(\mathrm{mod}11)$, the inverse is 5.

Multiply both sides by 5:

So $k=11m+5$ for some integer $m$.

Step 4: Find the general solution and values less than 200

Substitute back:

For $n<200$:

Verification:

Answer: The numbers are 49 and 148 (Option B)

To count how many times digit 7 appears in numbers from 1 to 100, we organize by position: units place and tens place.

Step 1: Count 7 in the units place

The digit 7 appears in the units place in: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97

Step 2: Count 7 in the tens place

The digit 7 appears in the tens place in: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79

Step 3: Account for overlaps

The number 77 contains the digit 7 twice (once in units place, once in tens place). We've counted it correctly in both steps above—no double-counting error since we're counting occurrences of the digit, not distinct numbers.

Step 4: Total count

Answer: The digit 7 appears 20 times in numbers from 1 to 100. (Option D)

Last digits of powers of 3: 3¹=3, 3²=9, 3³=27(7), 3⁴=81(1), 3⁵=243(3)...

Pattern: 3,9,7,1 repeats every 4. 2023 = 4×505 + 3, so 3^2023 has same last digit as 3³, which is 7.

First 20 multiples of 3: 3, 6, 9, ..., 60.

This is AP with a=3, d=3, n=20, l=60.

Sum = n(a+l)/2 = 20(3+60)/2 = 20×$\frac{63}{2}$ = 10×63 = 630.

100 = 2² × 5².

Sum of all divisors = (1+2+4)(1+5+25) = 7 × 31 = 217.

Sum excluding 100 = 217 - 100 = 117.

Using Legendre's formula: ⌊$\frac{27}{3}$⌋ + ⌊$\frac{27}{9}$⌋ + ⌊$\frac{27}{27}$⌋ = 9 + 3 + 1 = 13.

LCM(7, 11, 13) = 1001. Next multiple is 1001. 1001 - 1000 = 1. Actually 1000 ÷ 1001 remainder = 1000. Need 1001 - 1000 = 1. Recheck: 1000 mod 1001 = 1000, so add 1 gives 1001. Add 12 gives 1012 = 1001 + 11.

By Wilson's theorem, (p-1)! ≡ -1 (mod p) for prime p. So 16! ≡ -1 (mod 17). 16! = 13! × 14 × 15 × 16. -1 ≡ 13! × 14 × 15 × 16 (mod 17).

To find the sum of all odd divisors of 120, first extract the odd part by removing all factors of 2, then find the sum of divisors of that odd part.

Step 1: Prime factorization of 120

Step 2: Identify the odd part

The odd divisors of 120 come only from the odd prime factors (ignoring all powers of 2):

Step 3: Find all divisors of 15

The divisors of 15 are: $1,3,5,15$

These are exactly the odd divisors of 120 (since multiplying any of these by powers of 2 gives even divisors).

Step 4: Calculate the sum

Alternatively, using the divisor sum formula for $3^1\times 5^1$:

**Answer: The sum of all odd divisors of 120 is $24$ (Option C)**

Using Legendre's formula: ⌊$\frac{100}{5}$⌋ + ⌊$\frac{100}{25}$⌋ + ⌊$\frac{100}{125}$⌋ = 20 + 4 + 0 = 24

Let numbers be a and b. a + b = 15 and a² + b² = 117. We know (a+b)² = a² + 2ab + b². So 225 = 117 + 2ab, thus 2ab = 108, ab = 54

We need LCM(6, 8, 9). 6 = 2 × 3, 8 = 2³, 9 = 3². LCM = 2³ × 3² = 8 × 9 = 72. Smallest multiple of 72 greater than 100: 72 × 2 = 144.

Divisors of 20: 1, 2, 4, 5, 10, 20. Proper divisors (excluding 20): 1, 2, 4, 5, 10. Sum = 1 + 2 + 4 + 5 + 10 = 22, not 48.

Using Legendre's formula: floor($\frac{100}{5}$) + floor($\frac{100}{25}$) + floor($\frac{100}{125}$) = 20 + 4 + 0 = 24.

By Fermat's Little Theorem, since 13 is prime and gcd(5,13)=1, we have 5^12 ≡ 1 (mod 13). 100 = 12×8 + 4. So 5^100 ≡ 5^4 (mod 13). 5^4 = 625 = 48×13 + 1 ≡ 1 (mod 13).