Excel - formula to find how many cells to sum of N - excel

Question

I want to know how many cells it take to sum N. Please see following example:
number | cells to sum of 100
100 | 1
50 | 2
20 | 3
25 | 4
15 | 4
90 | 2
10 | 2
See the last column, it find the min number of current cell + previous cells to sum of 100.
Is there a way to do so?
Thanks.

Answer 1

In B2, array formula**:
=IFERROR(1+ROWS(A$2:A2)-MATCH(100,MMULT(TRANSPOSE(A$2:A2),0+(ROW(A$2:A2)>=TRANSPOSE(ROW(A$2:A2)))),-1),"Not Possible")
Copy down as required.
Change the hard-coded threshold value (100 here) as required.
As way of an explanation as to the part:
MMULT(TRANSPOSE(A$2:A2),0+(ROW(A$2:A2)>=TRANSPOSE(ROW(A$2:A2))))
using the data provided and taking the version of the above from B5, i.e.:
MMULT(TRANSPOSE(A$2:A5),0+(ROW(A$2:A5)>=TRANSPOSE(ROW(A$2:A5))))
the first part of which, i.e.:
TRANSPOSE(A$2:A5)
returns:
{100,50,20,25}
and the second part of which, i.e.:
0+(ROW(A$2:A5)>=TRANSPOSE(ROW(A$2:A5)))
resolves to:
0+({2;3;4;5}>=TRANSPOSE({2;3;4;5}))
i.e.:
0+({2;3;4;5}>={2,3,4,5})
which is:
0+{TRUE,FALSE,FALSE,FALSE;TRUE,TRUE,FALSE,FALSE;TRUE,TRUE,TRUE,FALSE;TRUE,TRUE,TRUE,TRUE})
which is:
{1,0,0,0;1,1,0,0;1,1,1,0;1,1,1,1}
An understanding of matrix multiplication will tell us that:
MMULT(TRANSPOSE(A$2:A5),0+(ROW(A$2:A5)>=TRANSPOSE(ROW(A$2:A5))))
which is here:
MMULT({100,50,20,25},{1,0,0,0;1,1,0,0;1,1,1,0;1,1,1,1})
is:
{195,95,45,25}
i.e. an array whose four elements are equivalent to, respectively:
=SUM(A2:A5)
=SUM(A3:A5)
=SUM(A4:A5)
=SUM(A5:A5)
Regards
**Array formulas are not entered in the same way as 'standard' formulas. Instead of pressing just ENTER, you first hold down CTRL and SHIFT, and only then press ENTER. If you've done it correctly, you'll notice Excel puts curly brackets {} around the formula (though do not attempt to manually insert these yourself).

Answer 2

I did the first 3 with an excel formula:
D3>100
C4 is where your numbers start, so C4=100, C5=50 etc.
Formula is on D4, D5, D6 etc
On D4:
=IF(C4>=D3;1;"False")
On D5:
=IF(C5>=D3;1;IF(C5+C4>=D3;2;"Error"))
On D6:
=IF(C6>=D3;1;IF(C6+C5>=D3;2;IF(C6+C5+C4>=D4;3;"Error")))
You can keep doing this, just keep replacing "Error" with an longer/updated version of IF(C6+C5+C4>=D4;3.
I don't know if this is the best way, but this will achieve it.

Answer 3

One way to solve this is to create an NxN matrix of equations instead of just a column. An example picture is provided. Columns E through I are hidden. The last column on the right determines the number required
Theoretically, you can also hard code the equations if the number of rows needed to get to 100 is a known small number. For example, if the number of rows is always four or less, C8 would be =IFS(B8>=100,1,SUM(B7:B8)>=100,2,SUM(B6:B8)>=100,3,SUM(B5:B8)>=100,4). BTW, you'll run into sum boundary problems with this equation on the first, second, and third rows. Therefore, the first row will need to be =if(B8>=100,1,""), the second row would be =IFS(B9>=100,1,SUM(B8:B9)>100,2,TRUE,"") and so on.