Prime Pennies Piles

For this problem, it is key to remember that 3 odd numbers always sum to an odd number; as all primes except 2 are odd, we know that 2 MUST be one of the piles.

From there, one simply has to refer to a list of primes and add 2 to each one, starting from 97, until he/she finds a prime P which, when it is added to 2, has the property that 100 - (2+P) is prime. Since 100 - (79+2) = 19, which is prime, 79 is our answer.

It is trivial to note that, by parity, one of the numbers must be even and thus $2$ . So we want the other two piles to have a sum of $98$ . We test primes backward from $97$ .

For $97$ , we get $1$ , which is not prime. For $89$ , we get $9$ , which is not prime. For $83$ , we get $15$ , which is not prime. Finally, we arrive at $79$ , which gets us $19$ .

Our answer is $\boxed {79}$ .

because prime numbers are odd except 2, then the sum of 3 prime numbers equal to 100, so one of them must be 2.

the sum of 2 prime numbers are 98 using

try and error we found that 79 + 19 = 98

Let each of the 3 piles have penny amounts of a,b,c respectively. WLOG, let $a\ge b\ge c$ . Then $a+b+c=100$ . Consider the equation modulo 2. This means that $a+b+c \equiv 0 \pmod{2}$ . If $a,b,c$ were all odd primes, then $a+b+c \equiv 1 \pmod{2}$ , absurd. Then, either $a$ is an even prime, or 2, or all three were even primes. Of course the latter doesn't make sense as $2+2+2<100$ . Then, let $a=2$ . Then $b+c=98$ . We than try the biggest possible primes for c, in specific (97, 89, 83, 79). If $c=97, b=1$ , not a prime. If $c=89, b=9$ , not a prime. If $c=83, b=15$ , absurd. If $c=79, b=19$ , hence $\boxed{79}$ is our biggest solution. $\blacksquare$

All prime numbers except 2 are odd (O). The only way to sum three numbers to an even (E) total given only one E is, O+O+E. Since we are looking for the largest value we must start with the largest prime (P) below 100 and consider the solution to 100-2-P. We continue to work down through the primes below 100 until that solution itself is prime. The first one we reach is 79.

We need to find three primes $a, b, c$ where $a \le b \le c$ such that $100=a+b+c$ . Notice that if $a, b, c >2$ then $a+b+c$ is odd, thus $a=2$ . Now we need to find $b, c$ such that $b+c=98$ . Obviously $b, c \ne 2$ . By constructing a table, we acquire the maximum possible value of $c$ is $79$

Let's assume each pile contain more than 2 pennies. Since all prime number larger than 2 are odd, their sum won't equal 100, contradiction. So,there is a pile which contain 2 pennies. The next task is to find 2 prime number whose sum is 98, which is 79 and 19.

When we take 79 we are left with 21 (sum of other two numbers) only 2 possibilities of primes is 19 and 2.

2+19+71=100