When \((629)^{24}\) is divided by 21, find the remainder.

Question

Let the number \((22)^{2022}\) + \((2022)^{22}\) leave the remainder \( \alpha \) when divided by 3 and \( \beta \) when divided by 7. Then \( (\alpha^2 + \beta^2) \) is equal to:}

Answer 1

There is a faster route here that avoids splitting 21 into factors at all. Look at how close 629 sits to a multiple of 21 itself.

Note that \(21 \times 30 = 630\). So \(629 = 630 - 1\), which means:

\[ 629 \equiv -1 \pmod{21} \]

Now raise both sides to the power 24. Since the exponent is even, a negative base raised to it becomes positive:

\[ 629^{24} \equiv (-1)^{24} = 1 \pmod{21} \]

So dividing \(629^{24}\) by 21 leaves a remainder of 1. None of the other options, 2, 5 or 11, matches this direct check.

Let's summarize:

629 is just 1 less than 630, and 630 is a clean multiple of 21 (21 x 30).
So 629 behaves like -1 under mod 21, and -1 raised to the even power 24 gives 1.

The remainder when 629^24 is divided by 21 is 1.

Answer 2

There is a faster route here that avoids splitting 21 into factors at all. Look at how close 629 sits to a multiple of 21 itself.

Note that \(21 \times 30 = 630\). So \(629 = 630 - 1\), which means:

\[ 629 \equiv -1 \pmod{21} \]

Now raise both sides to the power 24. Since the exponent is even, a negative base raised to it becomes positive:

\[ 629^{24} \equiv (-1)^{24} = 1 \pmod{21} \]

So dividing \(629^{24}\) by 21 leaves a remainder of 1. None of the other options, 2, 5 or 11, matches this direct check.

Let's summarize:

629 is just 1 less than 630, and 630 is a clean multiple of 21 (21 x 30).
So 629 behaves like -1 under mod 21, and -1 raised to the even power 24 gives 1.

The remainder when 629^24 is divided by 21 is 1.

Over Number	Run Rate
N-2	8.00
N	7.43
N+2	8.11
N+4	8.45
N+6	8.08

When \((629)^{24}\) is divided by 21, find the remainder.

Show Hint

The Correct Option is A

Solution and Explanation

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