Quantitative Aptitude - MCQ Practice Questions

Number of divisors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.

20 = 2² × 5. Divisors are: 1, 2, 4, 5, 10, 20. Sum = 1 + 2 + 4 + 5 + 10 + 20 = 42.

For n ≡ 3 (mod 7): possible numbers are 3, 10, 17, 24, 31, 38, 45, 52, 59... For n ≡ 5 (mod 11): possible numbers are 5, 16, 27, 38, 49, 60... Common number is 58. Check: 58 = 7(8) + 2... Let me recheck: $\frac{58}{7}$ = 8 rem 2, not 3. Try 38: $\frac{38}{7}$ = 5 rem 3 ✓, $\frac{38}{11}$ = 3 rem 5 ✓. Answer is A=38.

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.

To find the remainder when $2^{100}$ is divided by 7, we use Fermat's Little Theorem, which states that if $p$ is prime and $\gcd (a,p)=1$, then $a^{p-1}\equiv 1(\mathrm{mod}p)$.

Step 1: Apply Fermat's Little Theorem

Since 7 is prime and $\gcd (2,7)=1$:

Step 2: Express the exponent in terms of 6

Divide 100 by 6:

Therefore:

Step 3: Simplify using the congruence

**Step 4: Calculate $2^4\mathrm{mod}7$

Therefore:

**Answer: The remainder is $2$ (Option B)**

Using Euclidean algorithm: 144 = 96×1 + 48, 96 = 48×2 + 0. Therefore HCF = 48. Alternatively, 144 = 2^4×3^2 and 96 = 2^5×3. HCF = 2^4×3 = 48.

We need LCM(6, 8). 6 = 2×3, 8 = 2³. LCM = 2³×3 = 24. Therefore, 24 is the smallest number divisible by both 6 and 8.

# Unit Digit of 7^2019

To find the unit digit of any power, we identify the cyclical pattern of unit digits for that base number.

Step 1: Find the Cyclical Pattern of Unit Digits for Powers of 7

The unit digit of powers of 7 repeats in a cycle. Let's calculate the first few powers:

The pattern repeats: 7, 9, 3, 1, 7, 9, 3, 1, ...

The cycle length is 4.

Step 2: Find the Position of 7^2019 in the Cycle

Divide the exponent 2019 by the cycle length 4 to find which position in the pattern it corresponds to:

Since the remainder is 3, we need the 3rd unit digit in our cycle pattern (7, 9, 3, 1).

The 3rd position corresponds to unit digit 3.

The unit digit of 7^2019 is 3.

Answer: (A) 3

Using the property: HCF × LCM = Product of two numbers. 6 × 60 = 12 × x. 360 = 12x. x = 30.

To find $3^5\mathrm{mod}11$, we calculate successive powers of 3 and reduce modulo 11 at each step to keep numbers manageable.

**Step 1: Calculate $3^2\mathrm{mod}11$

Since $9<11$, we have $3^2\equiv 9(\mathrm{mod}11)$

**Step 2: Calculate $3^4\mathrm{mod}11$

Dividing: $81=7\times 11+4$, so $3^4\equiv 4(\mathrm{mod}11)$

**Step 3: Calculate $3^5\mathrm{mod}11$

Step 4: Final reduction

**Answer: $3^5\equiv 1(\mathrm{mod}11)$ (Option A)**

HCF takes minimum power of each prime: HCF = 2^min(4,3) × 3^min(3,2) × 5^min(1,2) = 2^3 × 3^2 × 5.

To find how many numbers between 50 and 150 are divisible by 7, we identify the smallest and largest multiples of 7 in this range, then count them using the arithmetic sequence formula.

Step 1: Find the smallest multiple of 7 greater than 50

Divide 50 by 7:

The next whole number is 8, so the smallest multiple is:

Step 2: Find the largest multiple of 7 less than 150

Divide 150 by 7:

The largest whole number is 21, so the largest multiple is:

Step 3: Identify the arithmetic sequence

The multiples of 7 from 56 to 147 form an A.P. with:

Step 4: Use the A.P. formula to find the count

For an A.P., $a_n=a_1+(n-1)d$

Answer: There are 14 numbers between 50 and 150 divisible by 7 (Option A)

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).

Let numbers be 3k and 5k. Since gcd(3,5)=1, LCM = 3k×5k/1 = 15k. Given LCM = 150, so 15k = 150, k = 10. Numbers are 30 and 50.

2^50 = 1,125,899,906,842,624. Sum of digits = 1+1+2+5+8+9+9+9+0+6+8+4+2+6+2+4 = 76. (Note: This requires calculation; the answer provided may vary based on computation.)

Using divisibility rule for 11: alternating sum of digits must be divisible by 11. For 121: (1-2+1) = 0, which is divisible by 11. Verification: 121 ÷ 11 = 11.

Smallest 4-digit number is 1000. For divisibility by 18, number must be divisible by both 2 and 9. 1000 ÷ 18 = 55.55... Next: 1008 ÷ 18 = 56. So 1008 is the answer.

Using the division algorithm: Number = (Divisor × Quotient) + Remainder. Number = (15 × 23) + 8 = 345 + 8 = 353.