# Problem of the Week Problem C and Solution Just Sum Primes

## Problem

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 $$a+b+c+d$$.

## Solution

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 $$27$$.