Number and Polynomial Division

LCM of 12, 15, 20, 54 is 540. Adding remainder 8 gives 548. Check: 548÷12=45 remainder 8, 548÷15=36 remainder 8, 548÷20=27 remainder 8, 548÷54=10 remainder 8.

Let the number be N and divisor be D. We have: (1) N ≡ 24 (mod D), (2) 2N ≡ 11 (mod D). From (2), 2N - 11 is divisible by D. Since N = kD + 24, we have 2(kD + 24) - 11 = 2kD + 48 - 11 = 2kD + 37. For this to be divisible by D, 37 must be divisible by D. Therefore, D = 37. We also need D > 24 (since remainder must be less than divisor), which 37 satisfies.

Let the divisor be d. Then: 242 = a×d + 8, so 242 - 8 = 234 = a×d. 698 = b×d + 9, so 698 - 9 = 689 = b×d. Also: 242 + 698 = 940 = c×d + 4, so 940 - 4 = 936 = c×d. The divisor d must divide all three: 234, 689, and 936. GCD of 234 and 936: 234 divides 936 (936/234 = 4), so GCD = 234. Factors of 234: 1, 2, 3, 6, 9, 13, 18, 26, 39, 78, 117, 234. Since d > remainders (8, 9), and d divides 689, we test: 689/13 = 53 exactly. So d = 13.

A number is divisible by 8 if its last three digits are divisible by 8. The sequence ends in ...7374. 374 divided by 8 is 46 with a remainder of 6 (8*46=368; 374-368=6).

Use Euler's theorem: 7 and 48 are not coprime (gcd=7), but observe the pattern: 7^1 ≡ 7 (mod 48), 7^2 = 49 ≡ 1 (mod 48), 7^3 = 7×1 ≡ 7 (mod 48), 7^4 = 7×7 = 49 ≡ 1 (mod 48). The pattern repeats every 2: odd exponents give remainder 7, even exponents give remainder 1. Since 123 is odd, the remainder is 7.

91+92+93+94+95+96+97+98 = 703, and 703/6 = 117 remainder 1, so remainder is 1, not 0. Alternatively, note that consecutive integers have remainders cycling 1,2,3,4,5,0. Summing eight consecutive numbers gives remainders summing to 0 (if sum of 6 consecutive numbers). Checking: 91+98=189 (divisible by 3), 92+97=189, 93+96=189, 94+95=189. Total 756, which is divisible by 6, so remainder should be 0. But wait, 91+92+93+94+95+96+97+98 = 703? No, that's wrong. Let me recalculate: (91+98)+(92+97)+(93+96)+(94+95) = 189+189+189+189 = 756. 756/6 = 126 exactly. So remainder is 0. The claimed answer is correct.

To find 2^1001 mod 1001, note that 1001 = 7 × 11 × 13. Using the Chinese Remainder Theorem: 2^1001 ≡ 4 (mod 7), 2^1001 ≡ 2 (mod 11), and 2^1001 ≡ 6 (mod 13). The unique number satisfying all three congruences is 123. Option A is correct.

The number N that leaves remainder 1 when divided by 2, 3 when divided by 4, 5 when divided by 6, and 7 when divided by 8 satisfies N+1 is divisible by 2, 4, 6, and 8. The LCM of 2,4,6,8 is 24. So N = 24k - 1. For N between 0 and 1000, k can be 1,2,...,42 (since 24*42-1=1007). Thus there are 41 numbers. Checking: 23, 47, 71, 95, ... up to 999.

Modulo 17: 19 ≡ 2, so 19^15 ≡ 2^15. 15 ≡ -2, so 15^19 ≡ (-2)^19 = -2^19. Expression = 2^15 - (-2^19) = 2^15 + 2^19 = 2^15(1+2^4) = 2^15(17) ≡ 0 (mod 17).

The MOD function in SQL returns the remainder of division. For example, MOD(10, 3) returns 1 because 10 divided by 3 equals 3 with remainder 1.

The MOD function in SQL returns the remainder of division. For example, MOD(10, 3) returns 1 because 10 divided by 3 equals 3 with remainder 1. This is a duplicate of question 139471.

MOD function returns the remainder when dividing two values. For example, MOD(10, 3) returns 1 because 10 divided by 3 is 3 with remainder 1. Option A describes QUOTIENT or division, not MOD.

MOD function returns the remainder after division. For example, MOD(10,3) = 1 because 10/3 = 3 remainder 1. Option A describes division with quotient (that's just / operator). Option B describes multiplication (that's * operator). Option D is false - MOD is a valid SQL function.

The MOD function returns the remainder when one number is divided by another. For example, MOD(10, 3) returns 1 because 10 divided by 3 equals 3 with remainder 1. Option A describes division with quotient (DIV function), option B describes multiplication, and option D is incorrect as MOD is a standard SQL function.

In IBM mainframe assembler, the Divide instruction uses an even-odd register pair. After division, the quotient is placed in the odd register and the remainder in the even register of the pair specified by the first operand.