I came across some problems and want to know what you think.

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    I came across some problems and want to know what you think. Here are the problems:

    How many pairs of line segments can you find in a cube that intersect each other? A. 12 B. 24 C. 36 D. 48 E. 60

    The sum of some consecutive odd numbers is 321. How many possible numbers are there at most?
    A. 1 B. 3 C. 107 D. 321 E. 642

    Given that A = 2^100, how many digits does A have? A. 30 B. 31 C. 32 D. 50 E. 100

    Nini has 10 oranges. If she can eat any number of oranges ( eating 0 orange is allowed ) each day, in how many ways can she eat all of them in 4 days? A. 286 B. 240 C. 1716 D.120 E. 720

  • M0★ M1★ M2★ M3★ M4 M5

    I'm not the best mathematician in the world but I'd like to point out some things:

    1. I'm not too sure of what you mean by pairs of line segments, because I can basically draw infinitely many line segments on each face of the square. I assume you are talking about the 12 sides as in line segments. Then, each side is always in contact with 4 other sides, so it's 12 choose 4, or 4 x 3 x 2 x 1 / 12 x 11 x 10 x 9 (possible outcomes / total outcomes). We don't really care about the ratio of possible to total outcomes though, we only want the possible outcomes, which is 4 x 3 x 2 x 1, which is 24, or B. Looking at the answer choices, that's most likely it, as if you counted all the overlaps, the answer would be far greater than any of the choices.

    2. As far as I know, that's impossible as any 2 odd numbers added up together can't form another odd number. Even if 0 was an odd number, it's not consecutive with 321.

    3. If you list out some of the squares of 2, you will find an interesting pattern: It takes 3 power to gain a new digit, 3 to get a new digit, 3 to get a new digit, and then 4 powers to get a new digit. 3 + 3 + 3 + 4 is 13, and 13 x 8 is 104. We're on to something! 13 in this case represents 4 powers of 2, so 4 x 8 minus 1 for the extra 4 in 104 makes 31 digits, or B.

    4. Assuming Nini always eats whole oranges (or no oranges), we have 11 options (she can eat no oranges) and 4 days in which we can use those options. But the choosing method doesn't work too well here for me because she can eat different amounts of oranges each day and the amount she eats can't be included in the next count. We can make a flow chart and see what happens.

    Eating 10 oranges, she has one path to go: eat no oranges for the rest 3 days. + 1 (possibility)

    Eating 9, she has one more orange left to consume on any one of the 3 days left. + 3

    Eating 8, she has 2 oranges left to eat on any one of the 3 days left or she could split them up 2 different ways, 1 on day 2 and 1 on day 3 or 1 on day 3 and 1 on day 4. + 5

    Eating 7, she has 3 oranges left for any one of the 3 days, again, or she could split the oranges 2 different ways, 2 on day 2 and 1 on day 3; 2 on day 3 and 1 on day 4; 1 on day 2 and 2 on day 3; or 1 on day 3 and 2 on day 4. +7

    See how we get 2 more possibilities for each extra orange? So we could use the choosing method, just a little differently. It would be 10 choose 4, but we count by 2 instead of 1, starting at 1, and we add instead of multiply.

    1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 20 (with eating no oranges on the first day, one path is repeated) = 120, or D. If overlaps of methods were allowed, again, the answers would be astronomical.

    Ok, so to sum it up:
    1. B
    2. N/A (???)
    3. B
    4. D (???)
    The ones with the question marks I'm not too sure about. I hope this helped. Please upvote because I spent 45 minutes on this thing.
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    @debbie @admins if you know how to solve this it would help, I'm still quite confused.

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    @The-Darkin-Blade upvoted

  • ADMIN M0★ M1 M5

    @bravekiwi Hi guys! 🙂 I forked this topic off to this post, since it wasn't exactly about the math question anymore 🙂

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    @debbie Thanks!! 😊

  • ADMIN M0★ M1 M5

    @smartlion You can also bring these questions to the Ask Math Anything live stream and ask Prof. Loh! That is probably the best way, since we (the moderators) are often too busy with other work to answer non-course-related math questions. 🙂