Number System and Digits

The unit digit of powers of 3 cycles in a pattern: 3, 9, 7, 1. Since 1244 is divisible by 4 (1244/4 = 311 with remainder 0), 3^1244 has the same unit digit as 3^4, which is 1.

Let the number be 10x + y where x is tens digit, y is units digit. Square: (10x+y)² = 100x² + 20xy + y². Dividing by half the number (5x + y/2) and adding 36 then dividing by 2 gives the reversed number 10y + x. Also x = 2(y-x). Testing option B: 46² = 2116, half of 46 = 23, 2116/23 = 92, 92+36 = 128, 128/2 = 64 ✓ And 4 = 2(6-4) = 4 ✓ The number is 46.

Using the numerals 1, 7, 7, 7, 7 to create 100: 177 - 77 = 100. This uses all five required digits exactly once. Other options like 177/77 ≈ 2.3, 717/77 ≈ 9.3, and 771/77 ≈ 10.0 are incorrect.

Even numbers are integers that are exactly divisible by 2. The sequence 2, 4, 6, 8 consists of positive even integers. Odd numbers are not divisible by 2, and real numbers include decimals and irrationals.

To find the unit digit, multiply the unit digits of each number: 4 * 8 * 7 * 3. 4 * 8 = 32 (unit 2); 2 * 7 = 14 (unit 4); 4 * 3 = 12 (unit 2). The final unit digit is 2.

Find unit digit of 2^100 - 3^50. Unit digits cycle: 2^n → 2,4,8,6 (period 4); 3^n → 3,9,7,1 (period 4). 100 mod 4 = 0, so 2^100 ends in 6. 50 mod 4 = 2, so 3^50 ends in 9. 6 - 9 = -3 ≡ 7 (mod 10). Option A (7) is correct.

The two-digit numbers that can be formed using the digits 4 and 5 (allowing repetition as implied by 'from 4 & 5') are 44, 45, 54, and 55. Their sum is 44 + 45 + 54 + 55 = 198. If repetition is not allowed, the sum is 45 + 54 = 99. The stored answer assumes no repetition.

For a number to be divisible by 4, its last two digits must be divisible by 4. From {1,2,3,4,5,6}, the pairs divisible by 4 are: 12, 16, 24, 32, 36, 52, 56, 64 (8 pairs). For each pair, we need to fill the first 3 positions from remaining 4 digits. Number of ways = 4P3 = 4 × 3 × 2 = 24. Total = 8 × 24 = 192.

Binary is a base-2 number system, meaning it uses only two digits: 0 and 1. Each digit represents a power of 2, making it fundamental to digital computing and the foundation of all modern computer systems.

We need a number consisting only of 3s that is divisible by 7. Testing 333, 3333, 33333, and 333333, we find $333333 / 7 = 47619$. This is the smallest such integer. Other options like 47719 or 48619 do not result in a product of all 3s when multiplied by 7.

The last digit of powers of 2 follows a repeating cycle of 4: 2, 4, 8, 6. For 2^18, we find the position in the cycle: 18 ÷ 4 = 4 remainder 2, so it corresponds to the second position in the cycle, which is 4. Therefore, 2^18 ends in 4.

Any power of 5 (where the exponent is at least 1) ends in the digit 5 (5, 25, 125, 625...). Therefore, 5^25 must end in 5. Options like 0, 10, or 50 are impossible for powers of 5.

Any power of a number ending in 1 will always end in 1 (1*1=1). Therefore, 11 raised to any positive integer power, including 11, will result in a number with 1 as the last digit.

Powers of 4 follow a cycle for the last digit: 4^1=4, 4^2=16, 4^3=64, 4^4=256. Odd powers end in 4 and even powers end in 6. Since 5 is odd, 4^5 ends in 4. (4^5 = 1024).