dont have book but need to do homework from chapter 5 of holt middle school course 1 math book?

Question

need to view lesson pages of book.0

Answer 1

http://go.hrw.com/hrw.nd/gohrw_rls1/pKeywordResults?keyword=MR7+HWHelp5

go here and see if this is what you need?

here is the main page to see if this is the book you need.....

http://go.hrw.com/gopages/ma/msm1_07.html

Answer 2

I did not want to answer this yet, but since the "binomial expansion" method is being attempted to prove the claim a tad bit too "trivially" I just want to clarify the non-obviousness of the problem: When you use binomial expansion, the sum of the coefficients for term 10^(i+j) that you get are NOT guaranteed to be digits in the resulting base 10 number! Think about it: Many such terms with large coefficients (as the resulting coefficients in the expansion are already product-coefficients) will be added together with no guarantees on the "spillovers" into higher digits. I did work more on this, steiner, both bottom-up and top-down, and while I did obtain the same result as you that any odd n satisfying the given properties would need to end in 001 (i.e. be congruent to 1 modulo 1000), the latter "continuation of cases" I got sort of exploded combinatorially. You probably tried these successive cases already, so I got to the sub-cases either t(250t +/- 1) results in a <0s&1s number base 10> or (2t+1)(125t + 62.5 +/- 0.5) results in a <0s&1s number base 10>. The "just keep testing the last digit of the product to eventually obtain a contradiction" thing doesn't seem to work since there are those powers of 5 that keep growing (well, perhaps you can use that fact to your advantage, as the growth must stop somewhere -- where?). I am a little bit pissed at you, Farful, and Low-Key for posting such interesting questions when I have so much damn work to do (midterm grading and two paper deadlines!). :P What I can see is that this is most definitely not your usual middle school problem! I doubt that I've been useful so far except to point out some unfortunate paths. My only other comment at the moment is this: What is so special about base 10? As Ana pointed out, of course the numbers do look like binary numbers, meaning that if the base were 2 the structure of the resulting squares would look more "random" as every x^2 would satisfy the property (it also means of course that for any max digit n, the total number of possible n satisfying the conditions is at most 2^n which is still too much for any exhaustive search). However, e.g. what happens when you ask the same question for base 3? I suppose the "oddness" constraint wouldn't hold anymore, so let's go one further and say: What happens when you ask the question in base 6? Or, as a different observation on divisors of 10, what happens in base 5? Good luck, steiner. :D ---- Dear Vikram: Why can't that even digit be zero, in which case what has been proved/disproved? ---- No problem, Vikram, it happens to all of us. The "middle school" part is kind of misleading too -- I felt really bad at first when I couldn't solve it immediately! I guess that's why steiner wrote "HARD" in all caps, hehe... Anyhoo, I'm done grading, partially done with paper-stuff... Off to sleep. :D G'night...

Answer 3

lesson pages of book 0???

Answer 4

Your local library may have an extra copy of the book. Call them and see.

Answer 5

just try calling 1 of your friends '_' Hope i could help!