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A Billion Bowling Pins
Week 5, 2026
All SolutionsPython - Inefficient but works method | BMC | Python Solutions
# meths or maths?
goals = [1_000_000, 1_000_000_000]
for goal in goals:
i = 0
while goal >= 0:
i += 1
goal -= i
print(i-1)
python | Peiran D | Python Solutions
//Paste your solution here if you want to share it publiclyrows = 0
number = 1
n = int(input())
while n > 0:
n -= number
number += 1
rows += 1
if n < 0:
rows -= 1
print(rows)
Python solution | AlfieJ | Python Solutions
n = int(input())
a = 0
b = 1
while True :
a += b
b += 1
print(a)
if a > n:
print(b-2)
break
elif a == n:
print(b-1)
break
Easier Python solution | AlfieJ | Python Solutions
n = int(input())
a = 0
b = 1
while a <= n :
a += b
b += 1
print(b-2)
python | JustinJ | Python Solutions
//Paste your solution here if you want to share it publicly
import math
x = int(input())
y = (-1 + math.sqrt(1 + 8*(x))) / 2
y= math.floor(y)
print(y)
python | JustinJ | Python Solutions
//Paste your solution here if you want to share it publicly
import math
x = int(input())
y = (-1 + math.sqrt(1 + 8*(x))) / 2
y= math.floor(y)
print(y)
Python | PaulZ | Python Solutions
import math
Pins = int(input("How many pins: "))
Rows = (-1 + math.sqrt(1 + 8*(Pins))) / 2
Rows= math.floor(Rows)
print(Rows)
Easiest but not efficient solution | MushyZ | Python Solutions
def calc_pin_rows(no_of_pins):
rows = 0
n = 1
if no_of_pins == 1:
return 1
while no_of_pins > 0:
if no_of_pins - n < 0:
break
rows += 1
no_of_pins -= n
n += 1
return rows
if __name__ == "__main__":
no_of_pins = 1000000000 #for billion
rows = calc_pin_rows(no_of_pins)
print(f"Number of rows needed for {no_of_pins} pins: {rows}")
