using python to get to a value with only known numbers
Question:
let’s say I have only have 2 known numbers, 1500 & 1000 and I want to figure out how many of each I need to add to get to 4000. what would be the most efficient way to code that?
Answers:
floor division, but with only two options, of course there still might be a remainder.
goal = 4000
a = goal // 1500
goal -= a * 1500
b = goal // 1000
goal -= b * 1000
results = {'balance': goal, '1500': a, '1000': b}
for k, v in results.items():
print(k, ':', v)
Here is one way you could do that.
#set variabes
N_1 = 1000
N_2 = 1500
N_3 = 4000
#set answer to how many times number 2 will go into number 3
Answer = N_3 / N_2
print (Answer)
# or you could do this to get number 1 will go into number 3
answer = N_3 / N_1
print (answer)
This will tell you how many times the numbers go into 4000 to change the numbers just set the variable to different values.
Solved this based on my situation… this is what worked for me but properly was luck based on the 2 numbers I was using.
goal = 4000
a = goal // 1500
goal -= a * 1500
b = goal // 1000
goal -= b * 1000
if goal == 0:
threetall = a
twotall = b
elif goal == 500:
goal += 1500
b = goal // 1000
threetall = a - 1
twotall = b
let’s say I have only have 2 known numbers, 1500 & 1000 and I want to figure out how many of each I need to add to get to 4000. what would be the most efficient way to code that?
floor division, but with only two options, of course there still might be a remainder.
goal = 4000
a = goal // 1500
goal -= a * 1500
b = goal // 1000
goal -= b * 1000
results = {'balance': goal, '1500': a, '1000': b}
for k, v in results.items():
print(k, ':', v)
Here is one way you could do that.
#set variabes
N_1 = 1000
N_2 = 1500
N_3 = 4000
#set answer to how many times number 2 will go into number 3
Answer = N_3 / N_2
print (Answer)
# or you could do this to get number 1 will go into number 3
answer = N_3 / N_1
print (answer)
This will tell you how many times the numbers go into 4000 to change the numbers just set the variable to different values.
Solved this based on my situation… this is what worked for me but properly was luck based on the 2 numbers I was using.
goal = 4000
a = goal // 1500
goal -= a * 1500
b = goal // 1000
goal -= b * 1000
if goal == 0:
threetall = a
twotall = b
elif goal == 500:
goal += 1500
b = goal // 1000
threetall = a - 1
twotall = b