#206 - Concealed Square
Find the unique positive integer whose square has the form 1_2_3_4_5_6_7_8_9_0, where each “_” is a single digit.
Short and sweet. The first thing to note is that the square ends in a 0. The only square numbers that end in 0 are multiples of 100, which means the final blank is a 0. We can now instead look for a square that looks like 1_2_3_4_5_6_7_8_9.
We want a square that ends in 9, which means the square root of this number ends in a 3 or 7 (since $3^2=9$ and $7^2=49$, both end in 9s). We can brute force this problem.
The largest possible number is where only 9s are filled in i.e. 19293949596979899.The largest square less than this is $138\,902\,662^2=19\,293\,949\,510\,686\,244$. We start with 138902657, since we want a number ending in 3 or 7. Then we alternatively subtract 4 and 6 to keep the last digit 3 or 7. A short function to test if the square follows the pattern is also needed.
# file: "problem206.py"
def isValid(n):
# Only need to check until the "8"
# because our number ends in "900".
for k in range(1, 9):
if int(str(n)[2 * (k - 1)]) != k:
return False
return True
# Start with the root of the largest number
# 19293949596979899 -> 138,902,662
n = 1389026657
subtracting = 4
while not isValid(n ** 2):
n -= subtracting
# Alternate subtracting 4 and 6
# to get numbers ending in 3 and 7
if subtracting == 4:
subtracting = 6
else:
subtracting = 4
print(f'{n * 10}^2 = {n * n * 100}')
Running the loop, we get
1389019170^2 = 1929374254627488900
0.0008701999999999877 seconds.
Thus, our unique number whose square follows the pattern is 1389019170. The time is very quick because the answer is fairly close to the upper limit.
