Problem of the week, 161

A former colleague Bill Wardlaw (March 3, 1936-January 2, 2013) used to create a “Problem of the Week” for his US Naval Academy students, giving a prize of a cookie if they could solve it. One of them is given below.

The residue of an integer n modulo an integer d > 1 is the remainder r left when n is divided by d. That is, if n = dq + r for integers q and r with 0 < r < d, we write $r \equiv n \pmod d$ for the residue of n modulo d. Show that the residue modulo 7 of a (large) integer n can be found by separating the integer into 3-digit blocks $n = b(s)b(s-1)\dots b(1)$.(Note that b(s) may have 1, 2, or 3 digits, but every other block must have exactly three digits.) Then the residue modulo 7 of n is the same as the residue modulo 7 of $b(1) - b(2) + b(3) - b(4) + \dots \pm b(s)$. For example,
$n = 25,379,885,124,961,154,398,521,655 \pmod 7$
$\equiv 655 - 521 + 398 - 154 + 961 - 124 + 885 - 379 + 25 \pmod 7$ $\equiv 1746 \pmod 7$ $\equiv 746 - 1 \pmod 7$ $\equiv 745 \pmod 7 \equiv 3 \pmod 7$.
Explain why this works and show that the same trick works for residues modulo 13.

Problem of the week, 137

A former colleague Bill Wardlaw (March 3, 1936-January 2, 2013) used to create a “Problem of the Week” for his US Naval Academy students, giving a prize of a cookie if they could solve it. One of them is given below.

Chain addition is a technique employed in cryptography for extending a short sequence of digits, called the seed to a longer sequence of pseudorandom digits. Quoting David Kahn (in Kahn on Codes, MacMillan, New York, 1983, p. 154), “the first two digits of the [seed] are added together modulo 10 [which means they are added and the carry is neglected] and the result placed at the end of the [sequence], then the second and third digits are added and the sum placed at the end, and so forth, using also the newly generated digits when the [seed] is exhausted, until the desired length is obtained”. Thus, the seed 3964 yields the sequence 3964250675632195… .

Periodic pattern

a. Show that this sequence eventually repeats itself.
b. Show that the sequence begins repeating itself with “3964”.
c. EXTRA CREDIT: How many digits are there before the first repetition of “3964”?

Problem of the week, 148

A former colleague Bill Wardlaw (March 3, 1936-January 2, 2013) used to create a “Problem of the Week” for his US Naval Academy students, giving a prize of a cookie if they could solve it. One of them is given below.

Suppose p and q are each monic polynomials of degree 4 with real coefficients and the intersection of their graphs is {(1, 3), (5, 21)}. If p(3) – q(3) = 20, what is the area enclosed by their graphs?

A particularly simple puzzles on round pegs

I just thought of this simple (at least I think it is simple) puzzle.

Consider a long loop of string and a number (at least 2) of round pegs of radius 1 inch each, parallel to each other. Drape the string around the pegs and pull the pegs so that the string is tight, as in the picture (which has only 3 pegs). Notice some sections of the string are straight and some are curved (shown in red in the picture).

Why is the total length of the curved sections equal to $2\pi$?

This is related to a pulley puzzle of Harry Langman, published in 1949.

Michael Reid’s Happy New Year Puzzle, 2012

Here is another puzzle from Michael Reid on NOBNET. (Michael also notes that if you find a solution (a,b,c,d,e,f) then (d,e,f,a,b,c) is another solution.)

Puzzle: Replace a, b, c, d, e, f with the first six prime numbers, in some order, to make the expression $a\cdot b^c + d \cdot e^f$ equal to the New Year.

Best wishes for a happy, healthy and prosperous New Year!

Michael Reid’s Digit operation puzzles

Michael Reid (http://www.math.ucf.edu/~reid/index.html) posted these questions on Nobnet recently. I’m reposting them here.

Consider two rules for modifying a positive integer:

i) replace a substring of its digits by the square of the number represented by the substring
ii) if a substring is a perfect cube, replace the substring by its cube root

Neither the substring being replaced, nor the string replacing it may have leading zeroes.

For example, starting with 123, we may square 23′ to obtain 1529. Then we can square 29′ to get 15841, then square 5′ to get 125841, and then we could take the cube root of 125′ to get 5841, and so on.

What is the smallest number we can obtain from the number 2011 with a sequence of these operations?

Now suppose the two operations are instead

iii) replace a substring by its cube
iv) replace a substring which is a square by its square root

What is the smallest number we can get with these operations if we start from 2011?

Here is another “digit operations” puzzle that I hope you will find a nice challenge.

Consider two rules for modifying a positive integer:

i) replace a substring of its digits by 7 times the number represented by the substring
ii) if a substring is a perfect 7th power, replace the substring by its 7th root

Neither the substring being replaced, nor the string replacing it may have leading zeroes.

For example, starting with 347, we may replace the substring 34′ by 238′ (multiplying by 7) to obtain 2387. Then we can multiply the substring 3′ to get 22187. Now we can take the 7th root of 2187′ to get 23, and so on.

What is the smallest number we can obtain from the number 2011 with a sequence of these operations?

Now suppose the two operations are instead

iii) if a substring is divisible by 7, we may divide the substring by 7
iv) replace a substring by its 7th power

(which are the reverses of operations i) and ii) above).

What is the smallest number we can get with these operations if we start from 2011?

God’s number for the Rubik’s cube in the face turn metric

This is an update to the older post.

The story is described well at cube20.org but the bottom line is that Tom Rokicki, Herbert Kociemba, Morley Davidson, and John Dethridge have established that every cube position can be solved in 20 face moves or less. The superflip is known to take 20 moves exactly (in 1995, Michael Reid established this), so 20 is best possible – or God’s number in the face turn algorithm. Google (and, earlier, Sony) contributed computer resources to do the needed massive computations. Congradulations to everyone who worked on this!

This would make a great documentary I think!