Answer
When \(N\) is divided by 2, 3, 5, and 7, the remainder is 1 in each case.
\(N = 211\) is a prime number.
Work Step by Step
We start with
\[
N = 2\cdot3\cdot5\cdot7 + 1 = 210 + 1 = 211.
\]
**Dividing by 2:**
Since \(210\) is divisible by 2, adding 1 gives
\[
211 \equiv 1 \pmod{2}.
\]
**Dividing by 3:**
Since \(210\) is divisible by 3, we have
\[
211 \equiv 1 \pmod{3}.
\]
**Dividing by 5:**
Similarly, \(210\) is divisible by 5, so
\[
211 \equiv 1 \pmod{5}.
\]
**Dividing by 7:**
Because \(210\) is divisible by 7,
\[
211 \equiv 1 \pmod{7}.
\]
**Is \(N\) prime?**
To determine whether \(211\) is prime, we check divisibility by all prime numbers less than or equal to \(\sqrt{211}\). Since \(\sqrt{211} \approx 14.5\), we need to test the primes \(2, 3, 5, 7, 11,\) and \(13\).
- \(211\) is not even, so not divisible by \(2\).
- \(211 \equiv 1 \pmod{3}\) (as shown above).
- \(211 \equiv 1 \pmod{5}\).
- \(211 \equiv 1 \pmod{7}\).
- For \(11\): \(11 \times 19 = 209\) and \(211 - 209 = 2\), so \(211\) is not divisible by \(11\).
- For \(13\): \(13 \times 16 = 208\) and \(211 - 208 = 3\), so \(211\) is not divisible by \(13\).
Since none of these primes divides \(211\), we conclude that \(211\) is prime.
