thanks for help
24x2y8xy=3x
3.) 105 = 1*3*5*7
3 digit number : 3, 5, 7. total permutations : 3! = 6
4 digit number : 1, 3, 5, 7. total permutations : 4! = 24
total : 6 + 24 = 30
310≡37(modK)
310−37≡273≡0(modK)
Factor of 273 : 1, 3, 7, 13, 21, 39, 91, 273
Largest possible two-digit positive integer K is 91