Paula, I am sure Francis is as Parisian as French Baguette. However, have you seen the size of those baguettes?
Consider the integer N=2^1999 (2 raised to the 1999 power).
Is there a positive integer multiple of N whose decimal representation does not contain the digit 0?
Mark:
2^1999
Since you're asking, I'll guess yes.
Your guesses are as good as the answers. (Well, almost)
Yes. Work from right to left. Start with N, and notice that its rightmost digit is not zero. Find the rightmost zero; say it's in the Kth position. Then add (10^K)*N (that is, N shifted over K positions) to change that digit without affecting any to the right of it. Now find the next zero, and again add (10^J)*N for some appropriate J. With each iteration, the rightmost zero moves further to the left. Continue until the rightmost zero is at least 1999 places to the left (or there are none). Now our integer is (10^1999)*A + B, where B has no zeros. Subtracting off (10^1999)*A = (2^1999)*(5^1999)*A, we are left with B, which is a multiple of (2^1999) with no zeros in its decimal representation.
We are playing baseball on an airless planet. A fly ball is hit directly towards an outfielder. The fielder can detect the changing angle that the ball makes with the horizon, but cannot directly judge changes in distance (apparent size of the ball), and there is no apparent lateral motion.
How does he decide whether to move forward or backward to catch the ball?
BASEBALL
Can he determine the rate of change of the angle that the ball makes with the horizon?
Does he know how far he is from home plate?
Is the batter on steroids? Was he created, or did he evolve? Is the president trying to keep his wife on life support?
Please let me know if any of the above questions are irrelevant, as that will keep me from heading in a wrong direction.
Good questions, and vital to formulate the answer.
If he is correctly positioned, the derivative of (tangent theta) with respect to time is constant. Put another way, there are constants c and d such that tan(theta)=c*t+d where t=time. This is because the ball's height (above the fielder) is a quadratic function of time, with value 0 when t=0 (i.e. when it reaches the fielder), and its horizontal distance from the fielder is a linear function of time, also achieving the value 0 when t=0. Dividing, we see that the slope is of the form c*t+d.
Suppose C is a positive real number such that for all positive integers N it is true that (N raised to the C power) is an integer. Does C have to be an integer?
Why or why not?
Mark:
N^C
I think C has to be an integer. If it is not, then I don't think 2^C is an integer.
The answer is that C must be an integer. I will assume that C is a non-integer that lies between 3 and 4. The proof generalizes but the notation gets harder. Pick an arbitrary large integer N. Consider the function f(x)=x**C, by which we denote "x raised to the C power". We will take the Taylor expansion of f(x) around the point N, and evaluate f(x) at the various points N+i as i=0,1,...,4. By hypothesis, since N+i is a positive integer, so is f(N+i). We will take a certain integer linear combination of the values f(N+i), concocted to make most of the terms of the Taylor expansion cancel out, and we will end up with a quantity which is supposed to be an integer, but instead lies strictly between 0 and 1.
The Taylor expansion tells us f(N+i) = f(N) + f'(N)i + f''(N)(i**2)/2 + f'''(N)(i**3)/6 + f''''(N)(i**4)/24 + f'''''(xi)(i**5)/120, where xi is some number between N and N+i. Now multiply the integers f(N+i) by (1, -4, 6, -4, 1), respectively, and sum. So we have: 1*f(N+0) - 4*f(N+1) + 6*f(N+2) - 4*f(N+3) + 1*f(N+4). On the one hand, we are multiplying integers by integers, so our resulting sum must be an integer. On the other hand, the sum comes out as 1*f(N) - 4*(f(N) + f'(N) + f''(N)/2 + f'''(N)/6 + f''''(N)/24 + f'''''(x1)/120) + 6*(f(N) + 2 f'(N) + 4 f''(N)/2 + 8 f'''(N)/6 + 16 f''''(N)/24 + 32 f'''''(x2)/120) - 4*(f(N) + 3 f'(N) + 9 f''(N)/2 + 27 f'''(N)/6 + 81 f''''(N)/24 + 243 f'''''(x3)/120) + 1*(f(N) + 4 f'(N) + 16 f''(N)/2 + 64 f'''(N)/6 + 256 f''''(N)/24 + 1024 f'''''(x4)/120). Now, many of the terms cancel. The coefficient of f(N) is (1-4+6-4+1)=0. The coefficient of f'(N) is (0-4*1+6*2-4*3+1*4)=(-4+12-12+4)=0. Similarly the coefficients of f''(N) and f'''(N) both vanish. The f''''(N) term has a coefficient of 1: (0 - 4*1 + 6*16 - 4*81 + 1*256)/24 = 24/24 = 1.
The error terms are all constant multiples of f'''''(xi). Now we know that f''''(N) = C*(C-1)*(C-2)*(C-3)*N**(C-4), and since C is between 3 and 4, we have that N**(C-4) grows smaller as N grows larger. For N large enough, f''''(N) is strictly between 0 and 1/2. (The coefficient C*(C-1)*(C-2)*(C-3) doesn't vanish because C is not an integer; this coefficient is some constant E independent of N.) The error term, involving the fifth derivative of f at various places, is bounded by some constant times N**(C-5). So the total sum is a nonzero constant times N**(C-4), plus an error term bounded by another constant times N**(C-5). For N large enough, this sum is bounded strictly away from 0, but smaller than 1 in absolute value. So we have an integer lying strictly between 0 and 1 (or between 0 and -1). By this contradiction we see that such a C cannot exist.
Let's borrow notation from LaTeX and let "^" denote exponentiation, so that x^2 means "x squared".
Consider the numbers 1233 = 12^2 + 33^2, and 990100 = 990^2 + 100^2.
Can you find an eight-digit number N with the same property, namely that if you break N into two four-digit numbers B and C, and add their squares, you recover N
(N really has to have eight digits; its leading digit is not zero. B is the number formed by the first our digits of N, and C is the number formed by the last four digits of N.)
I have decided to revamp the tax code in a major way. All residents will line up, left to right, with Sam standing at the far left at position 0 and the rest of us in positions 1, 2, 3, etc. On January 1 of each year, each resident will observe the net worth of the person on his left. On April 15, this resident pays that much in tax out of his own pocket. (So if Jack is at position 20 and Amanda at position 19, on April 15 Jack will pay the amount that Amanda had on January 1.)
"But what if I don't have that much money?"
Just run a deficit. Does that bother you?
"But what if my neighbor is running a deficit on January 1?"
If your neighbor was running a ten dollar deficit on January 1, then on April 15 you give the government a ten dollar deficit. That's the same as if the government pays you ten dollars.
Sam: "I don't have a neighbor, since I'm at position 0."
That's okay, Sam, you alone don't have to pay taxes. Just pretend your non-existent neighbor has zero net worth.
"What if people change places, drop out, come into the system...?"
None of that is allowed. Furthermore, no money ever changes hands except for payment of taxes. Okay, no further questions.
"Wait, wait. Why are we doing a US-centric April 15 tax day, if IBM is an international company?"
I said no further questions.
Sam starts out with one dollar, and the rest of you start with varying amounts. Some of you will start with no money, in particular the four people on the far right-hand end of the line will be broke. Some of you might even start with deficits.
Four years pass. At the end of those four years, Sam still has his one dollar, and the only other people with nonzero net worth are at positions q, r, s and t, with q < r < s < t. Here q,r,s,t are unknown integers, not assumed to be consecutive, and the answer to the problem will involve only the values q,r,s,t.
How much did the government collect in the first year (net)
(That is, collections minus payouts.) Guesses are most acceptable.
Are you saying,...Francis is like a baby in a topless bar, and I accidently started dancing, with said baby? 
Adapted from the 1986 Putnam Examination, problem A-6.