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111 = 37(1 + 1 + 1) 222 = 37(2 + 2 + 2) 333 = 37(3 + 3 + 3) 444 = 37(4 + 4 + 4) 555 = 37(5 + 5 + 5) 666 = 37(6 + 6 + 6) 777 = 37(7 + 7 + 7) 888 = 37(8 + 8 + 8) 999 = 37(9 + 9 + 9)
Christine Eaves, from the same school, found some interesting patterns:
190 = 19(1 + 9 + 0) 280 = 28(2 + 8 + 0) 370 = 37(3 + 7 + 0) 460 = 46(4 + 6 + 0) 570 = 57(5 + 7 + 0) 640 = 64(6 + 4 + 0) 730 = 73(7 + 3 + 0) 820 = 82(8 + 2 + 0) 910 = 91(9 + 1 + 0)
and another set:
198 = 11(1 + 9 + 8) 288 = 16(2 + 8 + 8) 378 = 21(3 + 7 + 8) 468 = 26(4 + 6 + 8) 558 = 31(5 + 5 + 8) 648 = 36(6 + 4 + 8) 738 = 41(7 + 3 + 8) 828 = 46(8 + 2 + 8) 918 = 51(9 + 1 + 8)
There are even easier cases if you go with the 100s:
100 = 100(1 + 0 + 0) 200 = 100(2 + 0 + 0) etc
Altogether there are 180 K- up three digit numbers, and you can gather them in sets sharing such patterns. Can you use algebra to prove that $K$ cannot be larger than 100? Using this fact you might like to write a computer program to list all the three digit K-up numbers and to verify that there are 180 such numbers.