The remainder when \((15^{23}+23^{23})\) is divided by 19, is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
A remainder depends only on the remainders of the numbers used: a number that leaves remainder \(r\) on division by 19, raised to a power, leaves the same remainder as \(r\) raised to that power. Here 23 is 4 more than 19, and 15 is 4 less than 19.
\(23=19+4\), so \(23^{23}\) leaves the same remainder as \(4^{23}\)
\(15=19-4\), so \(15^{23}\) leaves the same remainder as \((-4)^{23}=-4^{23}\), the power being odd
Sum: \(4^{23}-4^{23}=0\), so the remainder is 0
Tip: For an odd power \(n\), \(a^n+b^n\) is always divisible by \(a+b\). Here \(15+23=38=2\times19\), so the sum is divisible by 19 at once.