erikvm wrote:
Bunuel wrote:
SOLUTION
When 10 is divided by the positive integer n, the remainder is n - 4. Which of the following could be the value of n?
(A) 3
(B) 4
(C) 7
(D) 8
(E) 12
Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).
Original question says that when 10 is divided by the positive integer n, the remainder is n-4, so \(10=nq+(n-4)\) and also \(n-4\geq{0}\) or \(n\geq{4}\) (remainder must be non-negative).
\(10=nq+n-4\) --> \(14=n(q+1)\) --> as \(14=1*14=2*7\) and \(\geq{4}\) then --> \(n\) can be 7 or 14.
Answer: C.
I simply did it this way:
\(y = 10n + (n - 4)\)
And then I just plugged in values for n. So:
\(10(0) + (0-4) = -4\)
\(10(1) + (1-4) = 7\) <--my answer
Is this correct reasoning, or simply a lucky coincidence?
hi
erikvm,
here it is a lucky coincidence..
here you have found the value of y and not n...
\(10(1) + (1-4) = 7\)... here n=1 and y=7.... but the question asks us the value of n...
the equation that you have formed is wrong ..
it says
when 10 is divided by the positive integer n, the remainder is n-4.. so the equation will be 10=xn+n-4..
what we have to find is that " when 10 is divided by n, the remainder is n-4"
now substitue the values
(A) 3.... 10/3 rem=1, which is not equal to n-4=3-4=-1.... not correct
(B) 4.... 10/4 rem=2, which is not equal to n-4=4-4=0.... not correct
(C) 7.... 10/7 rem=3, which is equal to n-4=7-4=3.... correct
(D) 8.... 10/8 rem=2, which is not equal to n-4=8-4=2.... not correct
(E) 12... 10/12 rem=10, which is not equal to n-4=10-4=6.... not correct
hope it is clear..
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