1 2 3 4 5 6 7 8 9
123456789+987654321+123456789+987654321=100
1+2+3+4+5+6+7+ 8*9 = 100
987654321
3 and 9. And they divide into 123456789 whether or not you use divisibility rules!
1, 5, 9, 6, 3, 2, 4, 8, and 7 are.
The factors of 123456789 are 1, 3, 9, 3607, 3803, 10821, 11409, 32463, 34227, 13717421, 41152263 Only 3, 3607 and 3803 are prime factors.
Yes, 123456789 is a multiple of 3. To find out if it's a multiple of 3, you add the digits. In this case you would add 1+2+3+4+5+6+7+8+9=45. Now you take that 45 and divide by 3. If there is no remainder it is a multiple but if there is a remainder it is not a multiple. So in this case 45/3=15 with no remainder.
There are 64 subsets, and they are:{}, {A}, {1}, {2}, {3}, {4}, {5}, {A,1}, {A,2}, {A,3}, {A,4}, {A,5}, {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3, 5}, {4,5}, {A, 1, 2}, {A, 1, 3}, {A, 1, 4}, {A, 1, 5}, {A, 2, 3}, {A, 2, 4}, {A, 2, 5}, {A, 3, 4}, {A, 3, 5}, {A, 4, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, {A, 1, 2, 3}, {A, 1, 2, 4}, {A, 1, 2, 5}, {A, 1, 3, 4}, {A, 1, 3, 5}, {A, 1, 4, 5}, {A, 2, 3, 4}, {A, 2, 3, 5}, {A, 2, 4, 5}, {A, 3, 4, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5}, {A, 1, 2, 3, 4}, {A, 1, 2, 3, 5}, {A, 1, 2, 4, 5}, {A, 1, 3, 4, 5}, {A, 2, 3, 4, 5}, {1, 2, 3, 4, 5} {A, 1, 2, 3,,4, 5} .
3-5, 2-5, 3-5, 2-2, 5-2, 5-2, 3-3, 4-3, 4-4,
3-4, 3-4, 3-4, 4-3, 2-2 or 4-5, 4-5, 4-3, 3-3, 3-3
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 12 + 34 + 56 + 78 + 91 + 7894561230 + 123456789 = 8018018326