To Infinity, and Beyond!

Computer Science Level 3

Palindromes are numbers which are equal to their reverse. For example-

${121} = {121}$

Calculate

$\sum_{n}^{\infty} \frac{1}{n}$

where $n$ is a palindrome in base 10.

The answer is 3.37028.

4 solutions

Brock Brown
Jan 7, 2015

from fractions import Fraction as frac
def palindrome(number):
    s = str(number)
    return s == s[::-1]
total = 0
n = 1
infinity = 100000
while n <= infinity:
    if palindrome(n):
        total += frac(1, n)
    n += 1
print "Answer:", float(total)

This has to be a CS problem, I think because this is a fact found with the help of programming.

Kartik Sharma - 6 years, 5 months ago

Updated to CS

Agnishom Chattopadhyay - 6 years, 5 months ago

Trying from 4-5 days at last gave off , can't do without computer science - I don't know anything about it

U Z - 6 years, 4 months ago

That is the point exactly! School Education imparts the illussion in the students that every problem in the real world looks like a textbook problem and can be solved with analytical methods.

Agnishom Chattopadhyay - 6 years, 4 months ago

@Agnishom Chattopadhyay – I agree with you! It is so damn true.

Kartik Sharma - 6 years, 4 months ago

@Agnishom Chattopadhyay – I am extremely sorry! I won't post any problems like these. Sorry :(

Vatsalya Tandon - 6 years, 4 months ago

@Vatsalya Tandon – Please do not be sorry. I'm glad that you posted this problem. Post more of these! :)

I was not criticising you, I was criticising the school system which is throwing us into the darkness.

Agnishom Chattopadhyay - 6 years, 4 months ago

good coding..........

Md Mohibullah - 6 years, 4 months ago

Shouldn't it be infinity. It contains the harmonic series \sum_{n=1}^inf = inf ?

Amogh Kamath - 6 years, 4 months ago

It is very much a convergent series. Because of the very high rate of growth of the denominator, I would suppose.

http://en.wikipedia.org/wiki/Palindromic number#Sum of the reciprocals

Kp Govind - 6 years, 4 months ago

We can show convergence pretty quickly.

Consider the $2k+1$ digit numbers. The number of palindromes is $10^{k+1}$ . Each reciprocal is bounded above by $\frac{1}{ 10^{2k} }$ ,hence their sum is bounded above by $\frac{ 1}{ 10^{k-1} }$ . Hence, the sum of all reciprocals of palindromes with odd number of digits, is $\frac{10}{1 - \frac{1}{10}} = \frac{100}{9}$ . Similarly, for the $2k$ digit numbers, the number of palindromes is $10^k$ , and each reciprocal is bounded above by $\frac{1}{10^{2k-1} }$ . Hence their sum is bounded above by $\frac{1}{10^{k-1}}$ . Hence, the sum of all reciprocals of palindromes with even number of digits is $\frac{1}{1 - \frac{1}{10} } = \frac{10}{9}$ .

Hence, the sum is finite.

Calvin Lin Staff - 5 years, 8 months ago

Mehul Chaturvedi
Mar 14, 2015

Python:

def re(n):   #i'm making reverse function as re(n)
    return int(str(n)[::-1])
l=0
for m in range(1,999999):
    k=re(m)
    if k==m:
        l+=1.0/m
print 'answer is:',l

Did you learn Python at school ?

A Former Brilliant Member - 6 years, 3 months ago

No, @Azhaghu Roopesh M I learned it 3 weeks ago ,with the help of sir, @Chew-Seong Cheong .And still i'm am not perfect

Mehul Chaturvedi - 6 years, 2 months ago

Arulx Z
May 17, 2015

double sum = 0;
for(int n = 1; n < 100000000;n++)
    if(Integer.toString(n).equals(new StringBuilder(Integer.toString(n)).reverse().toString()))
        sum += 1d / (double) n;
System.out.println(sum);

OMG Java code is smaller than Python!

Is that so:

1	`reduce(lambda x,y: x + int(str(y)[::-1]==str(y))(1.0/y), xrange(1,10*7),0)`

Agnishom Chattopadhyay - 5 years, 8 months ago

Haha ok :)

Arulx Z - 5 years, 8 months ago

No. It's not. Less line doesn't make it smaller. Anyway, great job with the code.

Horisadi Afyama - 6 years ago

Chuck Jerian
Mar 26, 2015

sum=0.0

for x in range(1,10):
    #print(x)
    sum += 1.0 / x
print(sum)
xsum=sum
for max in  [100,1000,10000,1000000]:
    sum=xsum
    for o in (0,1):
        for i in range(1,max):
            if o == 0:
                eii = 1
            else:
                eii = 10
            for ii in range(0,eii):
                x=str(i)
                y=x[::-1]
                if o == 1:
                    m = str(ii)
                else:
                    m = ""
                z=x+m + y
                #print(o,i,ii," ",z)

                sum += (1.0 / int(z))
    print(sum)

To Infinity, and Beyond!

The answer is 3.37028.

4 solutions

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