Hands-on Experience

To explore the numbers One to Nine further you may like to try these extra puzzles. The first set of these go further into the topics of Chapter 9. They probe the inner life of numbers that lies beneath the surface of a calculator.

GENTLE: Check that 11 divides into 2¹⁰ − 1 = 1023, 3¹⁰ −1 = 59048…

MODERATE: Find the factors of 999999 and so show that 3, 7, 11, 13, 37 are the only primes p such that the decimals for 1/p recur in cycles of 1, 2, 3 or 6. Which primes p have decimals repeating in cycles of 4?

HARD: Which primes p have decimals for 1/p repeating in cycles of 7?

EASY: Check that the remainders that arise in working out 1/17 give the same sequence as the pseudo-random sequence in Chapter 8.

HARD: Show why 1/5 is 0.001100110011 … in binary notation. A calculator supplied as software on a computer may be working directly with binary numbers, and thus have the feature that it cannot store 1/5 exactly. Alternatively, it may be designed to work as if it was storing numbers in base-10. Power Macs of the 1990s did the former; the Mac OSX does the latter. Experiment to see what your computer does. Ask for 1/5 and then do repeated multiplication by 2 or 10 to find out what it is actually storing. How does it store 1/3 or 1/7?

TRICKY: The fact that the decimal for 1/13 breaks into two cycles of 6 is equivalent to the fact that 36 and 49 are congruent to 10 modulo 13, so that 10 appears in the diagonal of the modulo-13 multiplication table. Why does this show that 1/13 has a recurring full-length cycle of 12 when expressed in base 6 or base 7?

DEADLY: 1/89 begins with 0.011235 … and this points to another appearance of the Fibonacci numbers. 1/89 is actually equal to 0/10 + 1/100 + 1/1000 + 2/10000 + 3/100000 + 5/1000000 + 8/10000000 + 13/100000000 … Show why this is true by multiplying this sum by 89 (hint: 89 = 100 − 10 − 1). Without doing a division sum, see how the decimal for 1/9899 begins.

TOUGH: For a quicker proof of Fermat’s Little Theorem, look at the pth row of the Pascal triangle from Chapter 6, for a prime p. All the entries other than the two 1’s are divisible by p. Why is this? So 2^p − 2 is divisible by p. As a slightly harder problem, find a way to extend this result to m^p- m.

As you may have noticed, not everything in this book has to be taken completely seriously. You have to judge this for yourself. Here is a last puzzle which poses a similar challenge, wrapping together logic, numbers, and some curiosities from life and art. The answers are all numbers from One to Nine which fit into a Sudoku. There are twenty-six clues, and you must solve most but not all of them to complete the Sudoku. Many of them make use of modular arithmetic. Some are quite straightforward, but some are as tricky as codebreaking. If you can solve them all, you can count yourself a true expert in counting from One to Nine!

A: Find the last digit of .

B: Take the sequence 123456789. How many permutations leave exactly four numbers unchanged? Use the penultimate digit of your answer.

C: Find the penultimate digit of the trillionth power of 2.

D: How many anagrams are there of ONETONINE? Use the sum of the middle three digits.

E: Find the 666th digit, after the decimal point, of 1/666.

F: Six women and six men are at a party. In how many ways can they be arranged into man-woman pairs? Use the second digit of your answer.

G: At the same party, in how many ways can they be arranged into man-man and woman-woman pairs? Use the last digit of your answer.

H: At the same party, in how many ways can the people be arranged into pairs irrespective of gender? Use the penultimate digit.

I: Take the sequence 123456789. How many permutations leave exactly three numbers unchanged? Use the third digit of your answer.

J: How many anagrams are there of ANAGRAM? Use the penultimate digit.

K: Assuming that the world was created in 4004 B.C.E., in which year was it 6000 years old? Use the last digit of the year.

L: Express the Mersenne prime 2 − 1 in base-eight notation, and use the first digit.

M: Find the penultimate digit of the trillionth Fibonacci number.

N: Find the minus-third Fibonacci number.

O: Add the two last digits of the Mersenne prime 2^32582657 − 1.

P: Find the last digit of .

Q: There is a sequence starting with the DVD number, followed by the cult number with a film about it, then the ninth Fibonacci number, then the Ultimate Answer to Life, the Universe and Everything, then the difference between 1956 and 2006. Find the next number in this sequence, and use its first digit.

R: Find a number D between 1 and 32040 such that 17 × D = 1 (modulo 32040). Use its middle digit.

S: In the RSA system, with N = 143, M = 60, E = 11, find C.

T: Find the highest common factor of 12345679 and 888888; use its first digit.

U: Find the next number in the sequence beginning 19, 104, 22, 227, 129, 193, 241; use its last digit.

V: Find the 65537th digit after the decimal point of 1/12345679.

W: Find the last digit of .

X: If the Time Traveller arrives on the first Friday of January 802701, and the Gregorian calendar is still in force, which day of the month is this? (January 1, 1900 was a Monday.)

Y: Find the next number in the sequence beginning 19, 104, 177, 22, 161, 232, 39; use the last digit.

Z: A two-digit Fibonacci number appears at the beginning of 1984. Use the second digit.

Notes, links, updates and answers to all problems are on the Web at http://www.cryptographic.co.uk/onetonine