Number Theory and Pattern Puzzles Quiz

The sequence contains perfect squares: 1 (1²), 4 (2²), 9 (3²), 16 (4²), 25 (5²), and 36 (6²). The number 23 is the only non-square number, making it the clear outlier in this pattern of square numbers.

This sequence consists primarily of perfect squares: 1 (1²), 4 (2²), 9 (3²), 16 (4²), 36 (6²), and 49 (7²). The number 20 is the only non-square number, standing out as the exception to the square number pattern.

The pattern is n² + 1 where n = 1,2,3,...: 1²+1=2, 2²+1=5, 3²+1=10, 4²+1=17, 5²+1=26, 6²+1=37, 7²+1=50. The 8th term should be 8²+1=65, not 64. Therefore 64 is the wrong number in the series.

Looking at digit sums: 10→1, 14→5, 16→7, 18→9, 21→3, 24→6, 26→8. The pattern is 1,5,7,9,6,8 - which are mostly +2 increments (1→3→5→7→9). The sequence should be 10,12,14,16,18,21,24,26. The number 21 breaks the +2 pattern in digit sums (should be around 20 for digit sum 2, or 30 for digit sum 3).

The series contains perfect squares: 16=4², 25=5², 36=6², 144=12², 196=14², 225=15². However 72 is not a perfect square (8.485² ≈ 72). All other numbers are perfect squares, making 72 the odd one out.

For half-yearly compounding at 5% annual rate: Each half-year earns 2.5% interest. First deposit (Jan 1): 1600 for 2 half-years = 1600 × 1.025² = 1681. Second deposit (July 1): 1600 for 1 half-year = 1600 × 1.025 = 1640. Total = 1681 + 1640 = 3321. Interest gained = 3321 - 3200 = 121. Therefore, option B is correct.

Let sum = P. Simple Interest for 2 years at 4%: 2 × 0.04 × P = 0.08P. Compound Interest for 2 years at 4%: P(1.04)² - P = P(1.0816 - 1) = 0.0816P. Difference = 0.0816P - 0.08P = 0.0016P = 1. Therefore, P = 1/0.0016 = 625. Option A is correct.

Using the compound interest formula: A = P(1 + r)^t, where Amount = 30000 + 4347 = 34347, P = 30000, r = 0.07. Then 34347 = 30000(1.07)^t, so 1.07^t = 34347/30000 = 1.1449. Checking: 1.07² = 1.1449. Therefore, t = 2 years. Option A is correct.

Maturity value = P(1 + r)^t = 8000(1.05)² = 8000 × 1.1025 = 8820. Therefore, Albert gets Rs. 8820 at maturity. Option C is correct.

At 20% compound interest, we need the smallest integer n where (1.20)^n > 2. Checking: n=1: 1.20, n=2: 1.44, n=3: 1.728, n=4: 2.0736. Since 2.0736 > 2, the answer is 4 years. Option B is correct.

Let the three consecutive odd integers be x, x+2, x+4. Then 3x = 2(x+4) + 3 = 2x + 8 + 3 = 2x + 11. Therefore, x = 11. The third integer = x + 4 = 11 + 4 = 15. Option D is correct.

Let the number be x. Then x + 17 = 60/x, giving x² + 17x - 60 = 0. Solving: x = (-17 ± √(289 + 240))/2 = (-17 ± √529)/2 = (-17 ± 23)/2. The positive solution is x = (-17 + 23)/2 = 6/2 = 3. Option A is correct.

Let the numbers be a and b. We have ab = 120 and a² + b² = 289. Then (a+b)² = a² + 2ab + b² = 289 + 2(120) = 289 + 240 = 529. Therefore, a + b = √529 = 23. Option B is correct.

Let the consecutive even numbers be x and x+2. The difference of squares is (x+2)² - x² = 4x + 4 = 84. Solving: 4x = 80, so x = 20. The numbers are 20 and 22. Their sum = 20 + 22 = 42. Option C is correct.

Let the digits be x, y, z. We know x+y+z=10 and y=x+z. Substituting y in the first equation gives x+(x+z)+z=10, so 2x+2z=10, meaning x+z=5, which means y=5. When reversed, the number increases by 99, so the middle digit must be 5 (100-1=99 works only with middle digit 5). The only option with middle digit 5 that satisfies the conditions is 253.

The unit digit of any power of a number depends only on the unit digit of the base. For 31, the unit digit is always 1, regardless of the exponent (1479). This is because 1 multiplied by itself any number of times remains 1. The pattern 31^1=31, 31^2=961, 31^3=29791 all end in 1.

To convert hexadecimal FC to decimal: F=15, C=12. Using positional notation: (15 × 16^1) + (12 × 16^0) = (15 × 16) + 12 = 240 + 12 = 252. In hex, each position represents a power of 16, starting from right (16^0, 16^1, 16^2, etc.).

To test divisibility by 11, calculate the alternating sum of digits: (1-4+3-1+4-3+1-4+3-1+4-3) = 0. Since the result is 0 (a multiple of 11), the number is divisible by 11. The pattern 143 repeats, and 143 itself equals 11×13, making any repetition divisible by 11.

Any power of a number ending in 6 will also end in 6. This is because 6 × 6 = 36 (ends in 6), and this pattern continues indefinitely. For 2166^2166, since the base ends in 6, the result must end in 6 regardless of the large exponent. This is a fundamental property of the digit 6 in base-10 arithmetic.

To find the last two digits of 19^244, we use modular arithmetic with mod 100. The powers of 19 follow a cycle: 19^1=19, 19^2=61, 19^3=79, 19^4=01 (mod 100), then repeats every 20 powers. Since 244 = 240 + 4 and 240 is a multiple of 20, we need 19^4 mod 100 = 21. This uses Euler's theorem and the cyclic nature of modular powers.