ISBNs...

Level pending

An ISBN ( International Stanadard Book Number) is a ten digit code which uniquely identifies a book. The first nine digits represent the Group, Publisher and Title of the book and the last digit is used to check whether ISBN is correct or not.

Each of the first nine digits of the code can take a value between 0 0 and 9 9 (inclusive). Sometimes it is necessary to make the last digit equal to ten; this is done by writing the last digit of the code as X X . To verify an ISBN, calculate 10 10 times the first digit, plus 9 9 times the second digit, plus 8 8 times the third and so on until we add 1 1 time the last digit. If the final number leaves no remainder when divided by 11 11 , the code is a valid ISBN.

For example:

02011003311 = 10 0 + 9 2 + 8 0 + 7 1 + 6 1 + 5 0 + 4 3 + 3 3 + 2 1 + 1 1 = 55 02011003311 = 10*0 + 9*2 + 8*0 + 7*1 + 6*1 + 5*0 + 4*3 + 3*3 + 2*1 + 1*1 = 55 Since 55 55 leaves no remainder when divisible by 11 11 , hence it is a valid ISBN.

The problem: Let ξ \xi be the prime factorization of the number of valid ISBN codes possible. Find the sum of powers of the prime factors in ξ \xi .

Details and assumptions

If the number of codes would have been 50 50 , then ξ \xi would have been 2 5 2 2 * 5^2 and hence the answer would have been 2 + 1 = 3 2+1=3 . However, 3 3 is not the answer. :(


The answer is 18.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Josh Rowley
Mar 7, 2014

Whatever the first 9 digits are there will be exactly 1 choice of the 10th digit which will cause the summation to be divisible by 11 since the final value can take all 11 residues modulo 11. So every selection of the first 9 digits gives rise to exactly 1 valid ISBN, and each digit can be 0 through 9, ie. 10 possibilities, so there are 1 0 9 = 2 9 × 5 9 10^9 = 2^9 \times 5^9 valid ISBNs, so the answer is 9 + 9 = 18 9 + 9 = \fbox{18}

Anuj Asher
Mar 7, 2014

There are 9 unique digits and the 10th digit will depend on the first nine digits, and each digit can have 10 values i.e. 0-9.

So the number of possible ISBNs is 1 0 9 10^{9} , which can be written as 2 9 × 5 9 2^{9}\times5^{9} . Adding the powers 9 + 9 = 18 9+9=\boxed{18} .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...