63%962+x=999

asked by guest
on Nov 21, 2024 at 8:04 pm



You asked:

Solve the equation \(63 \:\%\: 962 + x = 999\) for the variable \(x\).

MathBot Answer:

The solution is: \[x = 936\]


\[\begin{aligned}x + 63 &= 999\\x &= 936\end{aligned}\]


The notation \(m \% n\) denotes the remainder of the integer \(m\) divided by the integer \(n\). \(m \% n = r\) means that given integers \(m\) and \(n\), where \(n\) is non-zero, there exist integers \(q\) and \(r\), where \(\lvert r \rvert < \lvert n \rvert\), such that \(m = q n + r\). MathBot adopts the convention that the remainder \(r\) and the divisor \(n\) share the same sign. Alternate notations for \(m \% n\) are \(\operatorname{mod}(m, n)\) and \(m \bmod n\).