Number System and Digits

A SIM card's ICCID (Integrated Circuit Card Identifier) is a 19 or 20 digit number that uniquely identifies the SIM card. It consists of industry identifier, country code, issuer identifier, and individual card identification. 20 digits is the standard length.

An MSISDN (Mobile Station International Subscriber Directory Number) is the full international telephone number including country code and national significant number. The ITU-T E.164 standard limits MSISDN to maximum 15 digits. Country codes are 1-3 digits, followed by the subscriber number.

We have 5 distinct digits (0,1,2,8,9) and need 4-digit numbers without repetition. First digit cannot be 0: 4 choices. Then arrange remaining 3 from 4 available digits: 4P3 = 4×3×2 = 24. Total = 4×24 = 96. Option D is correct.

For the number 27, the sum of digits is 2 + 7 = 9. Multiplying the sum by 3 gives 9 * 3 = 27. This satisfies the condition. Other options like 36 (3+6=9, 9*3=27) or 30 (3+0=3, 3*3=9) do not match.

For a 4-digit number using {9,8,7,5,0}, the first digit cannot be 0 (4 choices). The second digit can be any of the remaining 4 (including 0), the third digit 3, and the fourth digit 2. Total = 4 * 4 * 3 * 2 = 96.

Let the two-digit number be 10a + b, where a and b are the digits. When interchanged, it becomes 10b + a. The difference is (10a+b) - (10b+a) = 9(a-b) = 36, so a-b = 4. The digit difference is 4.

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.

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.

Let the number be 10a + b where a is tens digit and b is units digit. Given a = b + 2. Also (10a + b)(a + b) = 252. Substituting a = b + 2: (10(b+2) + b)((b+2) + b) = 252, which simplifies to (11b + 20)(2b + 2) = 252. Testing b = 2 gives (42)(6) = 252, so the number is 42, which matches the condition that tens digit (4) exceeds units digit (2) by 2.

Let the number be 10a + b, where a and b are its digits. The condition is 10a + b = 3(a + b), which simplifies to 7a = 2b. Since a and b are digits from 0-9, the only integer solution is a = 2, b = 7. Checking: 27 = 3 × (2 + 7) = 3 × 9 = 27, which satisfies the condition.

Hexadecimal needs 4 bits per digit (16 = 2⁴). For 12 hex digits, need bits such that ceil(bits/4) ≥ 12. At 44 bits: 44/4 = 11 hex digits exactly. At 45 bits: 45/4 = 11.25, ceil = 12 hex digits. Minimum is 45 binary digits.