Triple Digit Hunting

Computer Science Level 3

$\large 1234567891011121314\ldots 9989991000$

The number above shows a concatenating of the first 1000 positive integers in ascending order. How many 3-digit substrings consist of the same digit?

Bonus : Generalize for $1234567891011121314\ldots 10^n.$

The answer is 36.

2 solutions

Andrew Ellinor
Mar 3, 2016

Sometimes, I can Python

biglist = [n for n in range(1001)]
bigstring = ' '
count = 0

for k in biglist:
   bigstring = bigstring + str(k)

for i in range(len(bigstring) - 2): 
    if bigstring[i] == bigstring[i + 1] == bigstring[i + 2]:
    count = count + 1

print count

For my generalization, I found that the number of 3-digit substrings consisting of the same digit is $n\cdot10^{n-1} + 800\left(\frac{10^{n - 3} - 1}{9}\right) + 94 + n$ ... I think.

Equivalently, here's a two-liner version of your code.

1
2
3

>>> k=''.join([str(i) for i in range(1,1001)])
>>> print(sum(k[i]==k[i+1]==k[i+2] for i in range(len(k)-2)))
36

Your generalization, although has a typo (fix: replace all $n$ by $n-1$ ), is quite interesting.

Prasun Biswas - 5 years, 3 months ago

A two liner. :O

Just beautiful.

Andrew Ellinor - 5 years, 3 months ago

Tricky question! I used the sub-string '000', '111', '222' etc. to count the matches. What it does is, it counts the sub-string '0000' as 1 instead of 2. So I ended up at 20 instead of 36. I was about to post a report, when I just looked at your code and here I am liking your answer.

Siva Bathula - 5 years, 3 months ago

Sometimes , you can snake :3

Nihar Mahajan - 5 years, 3 months ago

Hjalmar Orellana Soto
Mar 24, 2016

a=""
b=1
while b<=1000:
    a+=str(b)
    b+=1
n=0
c=0
while n<len(a)-2:
    if a[n]==a[n+1] and a[n]==a[n+2]:
        c+=1
    n+=1
print c

Triple Digit Hunting

The answer is 36.

2 solutions

0 pending reports