**Goal:**

To find three different prime numbers **A, B, C** so that their sum **(A + B + C)** and their product **A*B*C** end with the digit **7**

For example,

3 + 17 + 17 = 3**7** 3 * 17 * 17 = 86**7**

7 + 13 + 17 = 3**7** 7 * 13 * 17 = 154**7**

Find other solutions.

All the prime numbers under 1000:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997

http://en.wikipedia.org/wiki/Prime_number

If the first term, A = 3, then B and C need to end with the digit 7:

Here are more examples:

3 + 7 + 37 = 47 3 * 7 * 37 = 777

3 + 17 + 37 = 57 3 * 17 * 37 = 1887

3 + 37 + 47 = 87 3 * 37 * 47 = 5217

3 + 47 + 67 = 117 3 * 47 * 67 = 9447

3 + 67 + 97 = 167 3 * 67 * 97 = 19497

3 + 97 + 107 = 207 3 * 97 * 107 = 31137

3 + 107 + 127 = 237 3 * 107 * 127 = 40767

3 + 127 + 137 = 267 3 * 127 * 137 = 52197

3 + 137 + 157 = 297 3 * 137 * 157 = 64527

3 + 157 + 167 = 327 3 * 157 * 167 = 78657

3 + 167 + 197 = 367 3 * 167 * 197 = 98697

3 + 197 + 227 = 427 3 * 197 * 227 = 134157

3 + 227 + 257 = 487 3 * 227 * 257 = 175017

3 + 257 + 277 = 537 3 * 257 * 277 = 213567

3 + 277 + 307 = 587 3 * 277 * 307 = 255117

Can A = B = 3 ?

What if A ≠ 3 ?

Well, thinking about it from a general point of view, here are some constraints I came up with:

–all three of the numbers must have ones digits that are odd (fairly obvious, but multiplying anything by 2 produces an even number)

–moreover, (this probably doesn’t cover all the cases), but a combination of primes ending in 3, 7, 7 (as in your example) will provide the ending digit 7 for both addition and multiplication. and , both of which end in 7.

–for addition, any combination of a pair whose ones add to ten and a 7 would do the trick (e.g., ) however, which does not end in 7.