jpr200012 wrote:

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

My strategy was to create lists below:

n = 3, 4, 7, 8, 12

n-4 = -1(becomes 9), 0, 3, 4, 8

n/10 = R? = 3, 4, 7, 8, 4

There is no match between n-4 and n/10's R.

The solution uses 14 = ..., but I don't understand how they are using 14. Should the question have said a multiple of one of these numbers?

Algebraic approach:

THEORY: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}

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.

Hope it's clear. - from this \(14=n(q+1)\) how did you get this as \(14=1*14=2*7\) I mean how did you figure out that 7 is n and q is 1 and not some other number combination yielding 14