Problem 1383

Solution:

The answer is $\mathbf{(C)}$. We observe that the only arithmetic progression satisfying the condition is 5, 11, 17, 23, 29. Indeed, the five numbers $a, b, c, d, e$, being in arithmetic progression with common difference 6, leave five different remainders upon division by 5: in particular one of them is a multiple of 5, and therefore equal to 5, since it is a prime number. Since $5 < 6$, we must have exactly $a = 5$, from which $b = 11$, $c = 17$, $d = 23$, $e = 29$.

It is then easily verified that statements (A), (B), (D), (E) are true, while (C) is false. Note moreover that options (A), (B), (E) can be excluded even without explicitly determining the numbers $a, b, c, d, e$. Indeed
$$a + b + c + d + e = a + (a + 6) + (a + 12) + (a + 18) + (a + 24) = 5a + 60,$$
which is a multiple of 5, and hence statement (A) is true.

Moreover, since $a, b, c, d, e$ are all prime numbers, we cannot have $a = 2$ or $a = 3$ (in that case we would have $b = 8$ or $b = 9$, which are not primes). We thus have $a \geq 5$, from which $b \geq 11 > 10 \Rightarrow abcde \geq bcde \geq b^{4} > 10^{4}$, so (B) is true.

Furthermore, since $a, b, c, d, e$ are distinct primes, no prime factor appears with exponent 2 or more in the factorization of $abcde$: it follows that the only perfect square dividing $abcde$ is 1, and hence (E) is true.