How many ways can a set of numbers sum to at least a given amount?
488357
Eleven: 1 3 7, 3 8, 3 3 5 = 3 ways\nTwelve; 1 3 3 5, 5 7, 1 3 8, 3 9 = 4 ways\nThirteen: 3 3 7, 1 3 9, 5 8, 1 5 7 = 4 ways\nFourteen: 3 3 8, 1 3 3 7, 5 9, 1 5 8 = 4 ways\nFifteen: 3 3 9, 1 3 3 8, 3 5 7, 1 5 9, 7 8 = 5 ways \nSixteen: 1 3 3 9, 1 3 5 7, 1 7 8, 7 9, 3 5 8 = 5 ways\nSeventeen: 8 9, 1 7 9, 3 5 9, 1 3 5 8 = 4 ways\nEighteen: 1 8 9, 1 3 5 9, 3 7 8, 3 3 5 7 = 4 ways\nNineteen: 1 3 7 8, 3 7 9, 3 3 5 8, 1 3 3 5 7 = 4 ways\nTwenty: 1 3 7 9, 3 8 9, 3 3 5 9, 5 7 8, 1 3 3 5 8 = 5 ways\nTwenty-one: 1 5 7 8, 5 7 9, 1 3 3 5 9, 1 3 8 9, 3 3 7 8 = 5 ways\nTwenty-two: 1 5 7 9, 3 3 7 9, 1 3 3 7 8, 5 8 9 = 4 ways\nTwenty-three: 3 3 8 9, 1 3 3 7 9, 3 5 7 8, 1 5 8 9 = 4 ways\nTwenty-four: 1 3 3 8 9, 7 8 9, 3 5 7 9, 1 3 5 7 8 = 4 ways\nTwenty-five: 1 7 8 9, 1 3 5 7 9 = 2 ways\nTwenty-six: 3 3 5 7 8, 1 3 5 8 9 = 2 ways\nTwenty-seven: 3 7 8 9, 3 3 5 7 9, 1 3 3 5 7 8 = 3 ways\nTwenty-eighr: 1 3 3 5 7 9, 1 3 7 8 9, 3 3 5 8 9 = 3 ways\nTwenty-nine: 1 3 3 5 8 9, 5 7 8 9 = 2 ways\nThirty: 3 3 7 8 9, 1 5 7 8 9 = 2 ways\nThirty-one: 1 3 3 7 8 9 = 1 way\nThirty-two: 3 5 7 8 9 = 1 way\nThirty-three: 1 3 5 7 8 9 = 1 way\nThirty-four: cannot be done = 0 ways\nThirty-five: 3 3 5 7 8 9 = 1 way\nThirty-six: 1 3 3 5 7 8 9 = 1 way\n\n78 ways in all\nSystematic searching is the only way I could find.
1092372
6\n1+(1*2)+3\n1+2+3\n3+3\n6\n\nuse order of opperations.  PEMDAS
942916
your question has a problem...............\nif the sum is 13, then the other number has to be -35, which will not meet your other condition
938348
There are many combinations. One of them is 1,2,3,4,5,16. Is there some specific property you wanted these six numbers to have?
806827
depends what your trying to sum from the polynomial... what are you trying to get, theres to many thing just to say sum of given polynomials
587245
If you have n events where the number of outcomes for the first event is p1 and the number of outcomes for the second event is p2 and so on up to the nth event (pn), then the number of total ways the event can occur is\n\n(p1)(p2)(p3)...(pn).
361589
Sum of all numbers 1 to 100:\nS = n[2a1 + (n - 1)d]/2\n= 100[2(1) + (100 - 1)]/2\n= 100(101)/2\n= 5050\n\nSum of all numbers 3 to 99, multiples of 3:\nS = n[2a1 + (n - 1)d]/2\n= 33[2(3) + (33 - 1)3]/2\n= 33(6 + 96)/2\n= 1683\n\nSum of all numbers 5 to 100, multiples of 5:\nS = n[2a1 + (n - 1)d]/2\n= 20[2(5) + (20 - 1)5]/2\n= 20(10 + 95)/2\n= 1050\n\nSum of all numbers 15 to 90, multiples of 15:\nS = n[2a1 + (n - 1)d]/2\n= 6[2(15) + (6 - 1)15]/2\n= 6(30 + 75)/2\n= 315\n\nSum of integers from 1 to 100 which are not divisible by 3 and 5:\nS = sum(1-100) - sum(3-99) - sum(5-100) + sum(15-90)\n= 5050 - 1683 - 1050 + 315\n= 2632\n\nWe have to add in the sum(15-90) because in removing sum(3-99) and sum(5-100), we're taking out multiples of 15 twice.  Adding sum(15-90) ensures that we only take out 15 once, 30 once, etc.
861916
I can't see the rest of your question, but even numbers can be multiples of three. Like 6 (2x3). They are not mutually exclusive.
952696
n + (n - 5) + (n + 6) \n3n + 1
961655
Yes, I can.  Quite simple actually.\n\nBut ask yourself if you know what these words mean:\n\n- integer\n- sum\n- difference\n- product  (think mathematics, not retail)\n- quotient
1025267
Here is how:\n\n6+6+3\n6+5+4\n5+5+5\n\nnow change the first two around until you have all the possiblities.
920768
One solution is\n0, 2, 4, 6, 8, 10, 12, 1, 3, 5, 7, 9, 11\nBut there is a huge variety of solutions\n\nSorry Chris I did not copy so here is another\n\n0, 11, 5, 2, 9, 3, 1, 4, 10, 7, 12, 6, 8
157677
This question cannnot be answered due to a simple fact that there are infinite amounts of numbers, therefore is no limit to a higher or a lower number.
810912
105 or any number that equals 105 time any multiple of 3,5 or 7
827303
a(n) = a(n-1) + sum of digits of a(n-1)  \n \n25, 32, 37, 47, 58, 71, 79\n\nNext is 79 + 16 = 95.
641014
give them an amount of money that is a multiple of 18 Ex. 36,54, 72, 90
110013
Well, all numbers would have to be represented equally, right?  If there are 27 combination I assume that 1, 1, 1 is included in these combintations, not just every combination that includes all 3 numbers...  However, if all three numbers are represented equally, each number should appear 27 times.  So the answer should be 27*9, or 243.