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Problem of the Week
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 \(9\).

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