The seventh POTW has been posted. Enjoy!
While at the Indiana math conference, I had the pleasure of seeing a short magic show by Caleb Wiles, who lives in Indianapolis and was apparently a math major at one time. I was impressed! It turns out he will at some point be appearing on Penn and Teller's television show Fool Us! So, here's a short video of him performing some elegant sleight of hand:
- Log in to post comments
200
The two factors of 9984 were very likely to be near 100. The difference between 10,000 and 9984 is 16, which makes us check the factors 100 +/- 4, before doing a full prime factoring.
96*104 = 9984, the difference is 8, and 8 is also the greatest common divisor, yielding 12 and 13 respectively. These last have to be consecutive numbers in order for the difference to also be the GCD.
There aren't other solutions. The full factoring is (2^8)*3*13. This means the GCD has to be a power of 2 greater than 4. We already verified 8. Trying 16 results in 3 and 13, which clearly aren't consecutive.
AM,
I used similar reasoning except I started by finding the prime factorization of 9984, ehich you correctly gave above as 2^8*3*13. It is clear that two numbers whose product equals this value must have a GCD equal to a power of two. The largest such value possible is clearly 16.
I then looked at pairs of factors of 9984. The pair that has the smallest difference is 96 and 104. The next smallest factor of 9984 is 78. 9984 = 78*128. The difference between this pair is clearly larger than 16, so this cannot be the pair in which the GCD equals the difference. All other pairs will have a larger difference and can therefore be ruled out.
A quick calculation easily shows that the GCD of 96 and 104 is indeed 8, which is equal to their difference. Therefore the sum is 200.
Like Sean, I found the prime factorization (2^8*3*13). I also noticed, as Matt did, that 9984 = 100^2 - 4^2, so I tried 96 = 2^5*3 as a factor. That leaves 104 = 2^3*13 as the other factor, and their GCD of 8 does in fact equal their difference. So the sum is 200.
I like to solve by computer, the factors were 3,13,and 256.
Decompose to n=n1*n2 and m=m1*m2
Now elaborate n2=(1,2,4,8,16) m2=256/n2
then a nested loop m1=(1,3,13,39) n1=39/n1
So there are 20 factorizations to try out, I simply printed them all out, and only 96*104 satisfies the requirement.
The only clever mathish optimization was exploitation of symmetry (n and m can be interchanged) so the outer loop only needs five rather than nine members.
Even if one had to do this by hand, it would only take a few minutes.....
Question about POTW 8:
====================================
Let A be the set {10, 11, 12, . . . , 20}, and let B
be the set {21, 22, 23, . . . , 30}. Each element
of the first set is multiplied, in turn, by each
element of the second set. Find the sum of all
these products.
====================================
Does "in turn...by each" mean A1*B1+A2*B2+A3*B3+....+A10*B10
or
A1*(B1+B2+B3+...+B10)+A2*(B1+B2+B3+...+B10)+....+A10*(B1+B2+B3+...+B10)
??
Another Matt, I had the same question initially and went with the second choice.
@Matt: The clue to resolving the question is that set A has 11 members and set B has only 10. So it has to be the second choice. If both sets had the same number of elements, then it would indeed be ambiguous.