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A Billion Bowling Pins
Week 5, 2026
Python - 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)
using vectors and loops | Draw6 | C++ Solutions
//Draw6
#include <iostream>
#include <vector>
// not efficient at all - takes about 45 seconds to run
// Function to generate the first n triangular numbers
std::vector<long> generateTriangularNumbers(long numberOfTerms) {
std::vector<long> triangularNumbers;
for (long n = 1; n <= numberOfTerms; ++n) {
long triangularNumber = (n * (n + 1)) / 2;
triangularNumbers.push_back(triangularNumber);
}
return triangularNumbers;
}
int main()
{
std::cout << "Please stand by for outcome...\n";
long numberOfTerms = 1;
long numM, numB, finalM, finalB;
for(int i = 0; i < numberOfTerms; i++)
{
std::vector<long> triangularNumbersM = generateTriangularNumbers(numberOfTerms);
finalM = numberOfTerms;
for ( long numM : triangularNumbersM)
{
if(numM > 1000000)
{
numberOfTerms = 0;
}
}
numberOfTerms++;
}
for(int i = 0; i < numberOfTerms; i++)
{
std::vector<long> triangularNumbersB = generateTriangularNumbers(numberOfTerms);
finalB = numberOfTerms;
for ( long numB : triangularNumbersB)
{
if(numB > 1000000000)
{
numberOfTerms = 0;
}
}
numberOfTerms++;
}
std::cout << "Number of rows with a million pins: " << finalM -1 << "\n";
std::cout << "Number of rows with a billion pins: " << finalB -1 << "\n";
return 0;
}
Loop, increment, subtract | greenya | Odin Solutions
package main
import "core:fmt"
main :: proc () {
for i in ([?] int {
0, 1, 2, 3, 4, 5, 6, 7, 8, 10,
20, 30, 40, 50, 60, 70, 80, 90, 100,
1e3, 1e4, 1e5, 1e6, 1e9, 1e12, 1e15, 1e18,
}) {
rows := full_rows_count(i)
fmt.printfln("%i\t%i", i, rows)
}
}
full_rows_count :: proc (pins: int) -> int {
i := 0
p := pins
for p >= 0 {
i += 1
p -= i
}
return i - 1
}
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)
Formula solution | AlfieJ | C++ Solutions
#include <iostream>
#include <cmath>
using namespace std;
int main() {
int n;
cin >> n;
int k = (sqrt(8 * n + 1) - 1) / 2;
cout << k << endl;
}
triangular numbers | AbsolutelyArsenal | Pseudocode Solutions
DECLARE Counter : INTEGER
DECLARE Triangle : INTEGER
Counter ← 1
Triangle ← (1/2)*(Counter)*(Counter + 1)
WHILE Triangle < 1000000
Counter ← Counter + 1
Triangle ← (1/2)*(Counter)*(Counter + 1)
ENDWHILE
OUTPUT Counter-1
WHILE Triangle < 1000000000
Counter ← Counter + 1
Triangle ← (1/2)*(Counter)*(Counter + 1)
ENDWHILE
OUTPUT Counter-1
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)
Pseudocode A Billion Bowling Pins | MarsKing | Pseudocode Solutions
FUNCTION HowManyRows(Pins:INTEGER) RETURNS INTEGER
DECLARE RemaningPins,Row:INTEGER
RemaningPins ← Pins
Row ← 0
REPEAT
Row ← Row + 1
RemaningPins ← RemaningPins - Row
UNTIL RemaningPins <= 0
IF RemaningPins < 0 THEN Row ← Row - 1 ENDIF
RETURN Row
ENDFUNCTION
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}")
