Divided by two numbers

[robertdseals]

I have a number and I want to see if it is divided by 10 and/or 8. 16 would because it can be divided by 8. 36 would because 2 10's and 2 8's. 17 would not because although you could get a 10 out of it, you wouldn't have enough left for an 8. Any thoughts?

 

[robertdseals]

All this does is identify even numbers. The numbers 12, 14, 22, 32 should all fail but are TRUE from your formula.

I think the question worded very poorly* but I think what he really wants is clear. The number conforms if it can be calculated as 8x + 10y where x and y are non-negative integers.

@robertdseals you have not answered my question about the possible range of numbers. The simplest method is the lookup table that I already provided but I can provide a VBA solution if that is not adequate.

*None of these numbers can be divided by 8 or 10.

No, and I'm not sure how I can make it any more clear. I wanted to know if a number can be divided by both 10, 8, or a combination of the two with no remainder. 12 would NOT be able to fit into this category.

 

[robertdseals]

It sounds like you are looking for a TRUE result if the number can be divided by either 8 or 10 and FALSE otherwise. If that is the case...

=OR(MOD(A1,8)=0,MOD(A1,10)=0)

I don't know if this would solve for 26 (with one 10 and two 8's).

 

[robertdseals]

This description does not indicate an actual result that can be shown in an Excel cell.

Are you looking for a Yes or No to indicate whether a given number can be created by summing only 10's and 8's?

If that is correct, this sounds like an iterative problem that can only be solved by Solver or VBA. I don't think this can be done with formulas. What kind of solution are you open to?

What is the range of possible numbers? It might actually be easier to generate all possible numbers that meet your criteria up to your limit.

$scratch.xlsm
	A	B	C	D	E	F	G	H	I	J	K	L
1		Number of 10's
2	Number of 8's	0	1	2	3	4	5	6	7	8	9	10
3	0	0	10	20	30	40	50	60	70	80	90	100
4	1	8	18	28	38	48	58	68	78	88	98	108
5	2	16	26	36	46	56	66	76	86	96	106	116
6	3	24	34	44	54	64	74	84	94	104	114	124
7	4	32	42	52	62	72	82	92	102	112	122	132
8	5	40	50	60	70	80	90	100	110	120	130	140
9	6	48	58	68	78	88	98	108	118	128	138	148
10	7	56	66	76	86	96	106	116	126	136	146	156
11	8	64	74	84	94	104	114	124	134	144	154	164
12	9	72	82	92	102	112	122	132	142	152	162	172
13	10	80	90	100	110	120	130	140	150	160	170	180
Sheet6

Cell Formulas
Range	Formula
C2:L2	C2	=B2+1
B3:L13	B3	=B$2*10+$A3*8
A4:A13	A4	=A3+1

Interesting solution! This might work. Thanks,

 

[robertdseals]

Your original description and your latest post, when taken together, are a little bit unclear to me. Let me ask this... what answer would you want for the numbers 18, 26, 28 and 34?

Yes, because 10 & 8, yes because 10 and two 8's, yes because two 10's and an 8, yes because 1 10 and three 8's.

 

[robertdseals]

Does this work for you

T202205a.xlsm
	A	B
1	18	TRUE
2	26	TRUE
3	28	TRUE
4	34	TRUE
5
1b

Cell Formulas
Range	Formula
B1:B4	B1	=OR(--RIGHT(A1,1)={0,2,4,6,8})

No, because it would give me true answers for 2,4,6,12. Niether which are divided by 10 and/or 8 (34 would because there are 3 8's and 1 10). I hope that makes sence.

 

[robertdseals]

All this does is identify even numbers. The numbers 12, 14, 22, 32 should all fail but are TRUE from your formula.

I think the question worded very poorly* but I think what he really wants is clear. The number conforms if it can be calculated as 8x + 10y where x and y are non-negative integers.

@robertdseals you have not answered my question about the possible range of numbers. The simplest method is the lookup table that I already provided but I can provide a VBA solution if that is not adequate.

*None of these numbers can be divided by 8 or 10.

EXACTLY! so, being divided by 8 and/or 10 would not leave any remainders.

 

[mikerickson]

If you have 10 in B1, 8 in C1 and 16 (or 36) in A2
Then in B2, put the CSE formula {=MATCH(0,MOD(A2-{0,1,2,3,4,5,6,7,8,9,10}*$B$1,$C$1),0)-1)} , which returns the number of 10's needed for the sum
in C2, =(A2-B2*$B$1)/$C$1 will return the number of 8's needed.

If the number in A2 cannot be expressed as a linear combination of B1 and C1, then they will return #NA.

 

[Fluff]

How about

++Fluff.xlsm
	A	B
1	1	FALSE
2	2	FALSE
3	3	FALSE
4	4	FALSE
5	5	FALSE
6	6	FALSE
7	7	FALSE
8	8	TRUE
9	9	FALSE
10	10	TRUE
11	11	FALSE
12	12	FALSE
13	13	FALSE
14	14	FALSE
15	15	FALSE
16	16	TRUE
17	17	FALSE
18	18	TRUE
19	19	FALSE
20	20	TRUE
21	21	FALSE
22	22	FALSE
23	23	FALSE
24	24	TRUE
25	25	FALSE
26	26	TRUE
27	27	FALSE
28	28	TRUE
29	29	FALSE
30	30	TRUE
31	31	FALSE
32	32	TRUE
33	33	FALSE
34	34	TRUE
35	35	FALSE
36	36	TRUE
37	37	FALSE
38	38	TRUE
39	39	FALSE
40	40	TRUE
Data

Cell Formulas
Range	Formula
B1:B40	B1	=MIN(MOD(A1-(ROW(INDIRECT("1:"&INT(A1/10)+1))-1)*10,8))=0
Press CTRL+SHIFT+ENTER to enter array formulas.

 

[KRice]

Or how 'bout this?

MrExcel_20220515.xlsx
	A	B	C	D	E
1	Value V	Divisor X	Divisor Y	aX+bY=V	a,b int.
2	18	10	8	TRUE
3	26	10	8	TRUE
4	28	10	8	TRUE
5	34	10	8	TRUE
6	36	10	8	TRUE
7	17	10	8	FALSE
8	12	10	8	FALSE
9	14	10	8	FALSE
10	22	10	8	FALSE
11	32	10	8	TRUE
Sheet6

Cell Formulas
Range	Formula
D2:D11	D2	=OR(MOD(A2-SEQUENCE(FLOOR(A2/B2,1)+1,,0)*B2,C2)=0)

 

[Fluff]

The OP's profile shows 2010 so won't have the Sequence function.