Question : A number $n$ when divided by 6, leaves a remainder of 3. What will be the remainder when $\left(n^2+5 n+8\right)$ is divided by 6?
Option 1: 1
Option 2: 3
Option 3: 5
Option 4: 2
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Correct Answer: 2
Solution : Euclid's Division Lemma, Dividend = Divisor × Quotient + Remainder Number = $n$ So, according to Euclid's Division Lemma $n= 6q+ 3$ where $n$ is dividend and $q$ is quotient. Also, $(n^2 + 5n + 8)$ when divided by 6 leaves a remainder, say $r$. $(n^2 + 5n + 8)$ = $(6q+3)^3 + 5(6q+3) +8$ = $36q^2+ 9+ 36q+ 30q+ 15+ 8$ = $36q^2+ 66q+ 32$ Now, consider each of the terms above separately to see what the remainder is when they are divided by 6. $36q^2$ is divisible by 6 as 6 is divisible by 6. Hence, the remainder is 0. $66q$ is divisible by 6 as 66 is division by 6. Hence, the remainder is 0. 32 when divided by 6 leaves a remainder of 2. $\therefore$ Remainder for $(n^2 + 5n + 8)$ is 2. Hence, the correct answer is 2.
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Question : N is the largest two-digit number which, when divided by 3, 4, and 6, leaves the remainder of 1, 2, and 4, respectively. What is the remainder when N is divided by 5?
Question : A number, when divided by 6, leaves a remainder of 3. When the square of the same number is divided by 6, the remainder is:
Question : M is the largest 4-digit number which, when divided by 4, 5, 6, and 7, leaves the remainder as 2, 3, 4, and 5, respectively. What will be the remainder when M is divided by 9?
Question : A number when divided by 221, leaves a remainder of 30. If the same number is divided by 13, the remainder will be:
Question : Let $x$ be the least number which, when divided by 5, 6, 7, and 8, leaves a remainder of 3 in each case, but when divided by 9, leaves no remainder. The sum of the digits of $x$ is:
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