Find the remainder when \(3^{3}\) is divided by 82.
Answer: D
\[ 3^{3} = 27 \]
Since \( 27 < 82 \), the division gives quotient 0 and the whole number is the remainder:
\[ 27 \]
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Which of the following numbers is not divisible by 18?
Answer: D
A number is divisible by 18 if it is divisible by its co-prime factors, 2 and 9.
Option D, 65043, ends in 3, so it is an odd number and not divisible by 2. Therefore, it cannot be divisible by 18.
Let's check the others to be sure. A) 34056: Ends in 6 (div by 2). Sum is 18 (div by 9). So divisible by 18. B) 54036: Ends in 6 (div by 2). Sum is 18 (div by 9). So divisible by 18. C) 50436: Ends in 6 (div by 2). Sum is 18 (div by 9). So divisible by 18.
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What is the remainder when the sum \(1^3 + 2^3 + 3^3 + ... + 100^3\) is divided by 4?
Answer: A
The sum of the first n cubes is given by the formula \(S_n = [\frac{n(n+1)}{2}]^2\). Here, n=100. Sum = \([\frac{100(101)}{2}]^2 = (50 \times 101)^2 = (5050)^2\). We need to find the remainder of \(5050^2 \div 4\). First, find the remainder of \(5050 \div 4\). The number formed by the last two digits is 50. \(50 \div 4\) gives a remainder of 2. So, we need to find the remainder of \(2^2 \div 4 = 4 \div 4\). The remainder is 0.
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