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Bunuel
When \(n\) is divided by 6, the remainder is 3, and the quotient is \(q\). When 3n+2 is divided by 6, the remainder is 5, and the quotient in terms of \(q\), is

A. \(\frac{q}{3}\)
B. \(\frac{q}{3}+2 \)
C. \(3q\)
D. \(3q + 1\)
E. \(3q + 2\)

­
Let the number n = 6q + 3

When 3n+2 is divided by 6, the remainder is 5. Substituting n in this equation we get,

3*(6q + 3) + 2 = 18 q + 11

taking 6 as common , 18 q + 11 becomes 18q + 6+5 = 6*(3q+1) +5

hence the new quotient in terms of q is 3q+1

Option D
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Use the division formula
n = 6q + 3

Now plug into 3n + 2:
3n + 2 = 3(6q + 3) + 2
= 18q + 9 + 2
= 18q + 11

Now divide 18q + 11 by 6:
18q + 11 = 6*(3q + 1) + 5

So quotient = 3q + 1 and remainder = 5

Answer: D


Bunuel
When \(n\) is divided by 6, the remainder is 3, and the quotient is \(q\). When 3n+2 is divided by 6, the remainder is 5, and the quotient in terms of \(q\), is

A. \(\frac{q}{3}\)
B. \(\frac{q}{3}+2 \)
C. \(3q\)
D. \(3q + 1\)
E. \(3q + 2\)

­
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n = 6q+3

3n+2= 18q+11= 18q+6+5
(3n+2)/6= 6(3q+1)+5

Thus quotient is 3q+1

Answer is D
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