# Math Q&A Game

by Gokul43201
Tags: game, math
 Emeritus Sci Advisor PF Gold P: 11,155 A Q&A game is simple: One person asks a relevant question (it can be research, calculation, a curiosity, something off-the-top-of-the-head, anything ... as long as it's a math question) and other people try to answer. The person who posts the first correct answer (as recognized by s/he who asked the question) gets to ask the next question, and so on. Let me start this off with a simple number theory problem : What is the least number than must be multiplied to 100! (that's a factorial) to make it divisible by $12^{49}$ ? (throw in a brief -couple of lines or so- explanation with the answer)
 P: 617 1/(99!*50)
 Emeritus Sci Advisor PF Gold P: 11,155 Not correct. "Multiplying" by your number gives a "product" of 2 (=100!/50*99!), and 2 is not divisible by 12^{49}.
 P: 617 Math Q&A Game i think i read it backwards...
 Sci Advisor HW Helper P: 9,396 There are 50 factors of 100 divisibile by 2, 25 by 4, 12 by 8, 6 by 16, 3 by 32 1 by 64 so the power of two in 100! is 97. similiarly there are 33 div by 3, 11 by 9, 3 by 27 and 1 by 81 making 48 times 3 divides, so i guess 2*3^50 will do
Emeritus
PF Gold
P: 11,155
 Quote by matt grime There are 50 factors of 100 divisibile by 2, 25 by 4, 12 by 8, 6 by 16, 3 by 32 1 by 64 so the power of two in 100! is 97. similiarly there are 33 div by 3, 11 by 9, 3 by 27 and 1 by 81 making 48 times 3 divides, so i guess 2*3^50 will do
I'm not sure I follow the finish...

You've shown that 100! has 2^{97} * 3^{48}. And what happened after that ?

In any case : Next question is yours ...
 Sci Advisor HW Helper P: 9,396 Duh, do i feel stupid from typing too quickly whilst someone is talking to me. should have been 6. Willl think of a question later tonight.
 Sci Advisor HW Helper P: 9,396 Ok, here's one that I hope isn't too tricky. Let A be an nxn matrix over C, the complexes. Suppose that Tr(A^r)=0 for all r. Show A is nilpotent. All the other ones I could think of were too easily looked up.
HW Helper
P: 3,684
 Quote by Gokul43201 I'm not sure I follow the finish... You've shown that 100! has 2^{97} * 3^{48}. And what happened after that ?
$$12^{49}=2^{98}3^{49}$$, so $$100!=2^{97}3^{48}X$$ must be multiplied by $$6=2^{98-97}3^{49-48}$$ before $$12^{49}$$ will divide it.
Emeritus
PF Gold
P: 11,155
Yes, CRG : matt clarified this in his subsequent post. Thanks !

Back to matt's question :

 Quote by matt grime Let A be an nxn matrix over C, the complexes. Suppose that Tr(A^r)=0 for all r. Show A is nilpotent.
 Sci Advisor HW Helper P: 9,396 If there's no progress on this after a couple of days i'll post hints. perhaps people should post what they think they need to do. i like tyhe question cos it uses lots of bits from here there and everywhere.
 P: 127 Well it seems that first you need to show that such a matrix does not have n distinct eigenvectors and then show that not having n distinct eigenvectors implies nilpotents. The first question relates to the fact that if a matrix had a full set of Eigenvectors then diagonalization gives us that $$\sum \lambda_i^n =0$$ for all n>0. For the second half I would gather you need to consider the vectors not in the span of the Eigenvectors and consider where they may be mapped. hmmm... that may not be enough for the first half. You may need to also show that any vector that is a eigenvector is in fact in the kernel. Experimentally it appears like we are dealing with upper or lower triangular matrix if you just assume Tr(A^r) =0 for r=1..n. But that could just be maple not returning all possible answers which it sometimes does. That's what I've been thinking
 Emeritus Sci Advisor PF Gold P: 16,091 Hrm. If a matrix A is diagonalizable, then I claim that there exists an n such that all of the nonzero eigenvalues of An lie in the right half-plane. The requirement that Tr(An) = 0, forces all the eigenvalues to be 0, and thus A is zero... clearly nilpotent! So the trick, then, is when the matrix is not diagonalizable. Then again, this only works for complex valued matrices.
 Sci Advisor HW Helper P: 9,396 I did state the matrix was over C, though this is largely for conveniece. And, snobel, the matrix 0 has a full set of eigen values and is certainly nilpotent an satisfies the criterion.
 P: 127 Oops... of course that is the sole matrix with a full set of 0 eigenvalues and $$\lambda_i=0$$ is the sole set of solutions in C satisfying my condition.
P: 2,751
 Quote by Gokul43201 What is the least number than must be multiplied to 100! (that's a factorial) to make it divisible by $12^{49}$ ?
Sorry but the correct answer is the rational number 12^49 / 100! , it's smaller than the previous answer of 6 by a factor or approx 10^104.

Ok so here's my QA puzzle : Why is it that mathematicians are worse than the general layperson when it comes to not specifying that they require a whole number solution when that is the case. :p
 Sci Advisor HW Helper P: 9,396 If you're going to take that attitude, there is no answer; think negatives.