Splitting 61 into integer partitions - algorithm

Question

In a program I am doing, I have a vector called a = [5, 6, 7] and I have to split integer 61 into additive partitions using integer from this list. So one example would be
61 = 5 + 6 + 7 + 7 + 6 + 6 + 6 + 6 + 6 + 6
There are many ways to split this. I have to do this programmatically. One approach I found is as follows. I don't know if this always will give result. First I check if 61 is divisible by any number in the list. If it is, then I can just use that number to add many times (i.e. quotient) to get 61. In this case, 61 is a prime number. So this will fail. Next step is to
take first number in the list (in our case, 5) and subtract it from 61 and try to see if the answer is divisible by any member in the list. If it is, then we again found a way to do addition. In this case, subtracting 5 from 61 gives 56, which is divisible by 7 and our solution would be
61 = 5 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7
In this manner we continue down the list until we find some answer after subtraction which is divisible by a member in the list.
Now the given list to me, [5, 6, 7] is such that there exists an integer partition such that, we can get 61 from additions using that integer partition. So we won't have to worry whether a solution exists. So my approach seems very crude. I wonder if there is an efficient way to do this
using some algorithm from combinatorics. So my final answer should be a list
of numbers from the integer partition. So one possible answer would be
[5, 6, 7, 7, 6, 6, 6, 6, 6, 6]
thanks

Answer 1

The generic root of 61 is 7
Multiple of 7 nearing 61 will be 8 hence , subtracting 61-7*8 =gave 56,
Similarly multiple of 6 nearing 61 with a gap of 7 will then be 9,
subtracting 61-6*9 =gave 54,then result % remaining two from {5,7} to be satisfied should be zero
Getting intermediate sum and finding their generic root combined with displacement can give the answer.
Hope this helps. !! Happy to help further

Answer 2

I think I found a crude logic here. Let's say given list is [5, 6, 7] and number is 61. we need to find additive list such that total is 61. one such example is
61 = 5 + 6 + 7 + 7 + 6 + 6 + 6 + 6 + 6 + 6
another would be
61 = 5 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7
Our job is to get the numbers on the right side as a list. So one possible solution would be
[5, 6, 7, 7, 6, 6, 6, 6, 6, 6]
My algorithm is as follows. First subtract all list members from 61. So 61 - 5 - 6 - 7 = 43. So we get first three members from the list, which are 5, 6, 7. Now remainder here is 43. To get other members, we subtract each of 5, 6, 7 from 43 , one at a time and see if the answer is divisible by any of 5, 6, 7. So
43 - 5 = 38 -> not divisible by any [5, 6, 7]
43 - 6 = 37 -> not divisible by any [5, 6, 7]
43 - 7 = 36 -> divisible by 6 in [5, 6, 7]
and quotient is 6, which means we have to use 6 of 6's and 7 which we subtracted last from 43. So the list would be original list plus 7 and a
list of 6 with length 6. So one possible solution I found is
[5, 6, 7, 7, 6, 6, 6, 6, 6, 6]
And we can verify that the sum is 61. I wrote a program in R programming language. Here it is
get_me_list <- function(number, mylist){
rest <- number - sum(mylist)
flag = FALSE
for(i in seq_along(mylist)){
answer <- rest - mylist[i]
for(j in seq_along(mylist)) {
if( answer %% mylist[j] == 0){
repeat_factor <- answer / mylist[j]
number_to_repeat <- mylist[j]
pivot <- mylist[i]
flag = TRUE
break
}
}
if(flag){
break
}
}
final_list <- c(mylist, pivot, rep(number_to_repeat, repeat_factor))
final_list
}
So get_me_list function takes two inputs. number and my_list. In my case number = 61 and my_list = [5, 6, 7] . In R, a vector is written as
c(5, 6, 7) or 5:7 if its a sequence from 5 to 7. So the output I get is
c(5, 6, 7, 7, 6, 6, 6, 6, 6, 6)
which is a vector in R. I tried giving various values of my_list and compared the solution with manual solution which I calculated using pen and paper. I am getting correct answer with the above code. I don't know if this approach would be valid always. I am off course assuming that its possible to get the sum equal to number using only the members of the my_list.
Please comment on code.
thanks