Problem C and Solution

Just Sum Primes

A *prime number* is an integer greater than \(1\) that has only two positive divisors: \(1\) and itself. The number \(17\) is prime because its only positive divisors are \(1\) and \(17\).

The variables \(a\), \(b\), \(c\), and \(d\) represent four different prime numbers. If \(a\times b\times c\times d\) is equal to a three-digit number with a tens digit of \(1\) and a ones (units) digit of \(0\), determine all the possible values of \(a+b+c+d\).

Let \(e\) be the hundreds digit of the product \(a\times b\times c\times d\). In other words, \(a\times b\times c\times d=e10\).

Since \(e10\) ends in \(0\), it must be divisible by \(10\), which is the product of the two primes \(2\) and \(5\). That tells us that \(2\) and \(5\) must be two of the prime numbers \(a,~b,~c,\) and \(d\).

When \(e10\) is divided by \(10\), the quotient is \(e10\div 10=e1\). Since \(e10\) is a three-digit number, \(e\not=0\) because \(e10=010=10\) is not a three-digit number. Thus, the possibilities for \(e1\) are \(11\), \(21\), \(31\), \(41\), \(51\), \(61\), \(71\), \(81\), and \(91\).

The two-digit number \(e1\) must be the product of two distinct prime numbers, neither of which is \(2\) or \(5\). We can rule out any possibilities for \(e1\) that are prime, since these numbers would have only one prime factor. Therefore, we can rule out \(11,~ 31,~ 41,~ 61,\) and \(71\), which are all prime. The remaining possibilities for \(e\) are \(2\), \(5\), \(8\), and \(9\).

If \(e=2\), then the two-digit number would be \(21\), which has prime factors \(7\) and \(3\). The four prime factors of \(e10=210\) are \(2\), \(3\), \(5\), and \(7\), producing a sum of \(2+3+5+7=17\).

If \(e=5\), then the two-digit number would be \(51\), which has prime factors \(17\) and \(3\). The four prime factors of \(e10=510\) are \(2\), \(3\), \(5\), and \(17\), producing a sum of \(2+3+5+17=27\).

If \(e=8\), then the two-digit number would be \(81\), but we cannot write \(81\) as the product of two distinct prime numbers. Note that \(810=2\times 3\times 3\times 3\times 3\times 5\), which is the product of six prime numbers not all of which are distinct. Therefore, \(8\) is not a possible value for \(e\).

If \(e=9\), then the two-digit number would be \(91\), which has prime factors \(7\) and \(13\). The four prime factors of \(e10=910\) are \(2\), \(5\), \(7\), and \(13\), producing a sum of \(2+5+7+13=27\). However, we already have the sum \(27\).

Since there are no other possible cases to consider, the only possible values of \(a+b+c+d\) are \(17\) and \(27\).