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    Can someone please confirm??

    "Ask Math Anything" Chat Group
    desolate101 eeveelution
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    • Desolate_101D
      Desolate_101 M0★ M1★ M2 M3 M4★
      last edited by Desolate_101

      75c01ea6-21ae-47b4-a222-a71a775dc74b-image.png

      how many rectangles are in here?
      This is on a book, and the book's answer is 73, but I count 76.
      Can anyone confirm either answer?

      Hello, you are now looking at a book worm who likes animations about stickman, and is very nearly bad at math.
      If you wanna see my animations, search "AshFang1130" on BilliBilli(哔哩哔哩)
      P.S. I LOVE WARRIOR CATS!!!!!!!!
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      • Desolate_101D
        Desolate_101 M0★ M1★ M2 M3 M4★
        last edited by Desolate_101

        Also, what is the last digit of 1^15 + 2^15 + 3^15 +......+ 2011^15 + 2012^15 ?

        Hello, you are now looking at a book worm who likes animations about stickman, and is very nearly bad at math.
        If you wanna see my animations, search "AshFang1130" on BilliBilli(哔哩哔哩)
        P.S. I LOVE WARRIOR CATS!!!!!!!!
        P.P.S I also play GachaClub

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        • Desolate_101D
          Desolate_101 M0★ M1★ M2 M3 M4★
          last edited by

          PS. for the first question, ignore the stuff I drew, and also, I can assure everyone that the books answer is usually wrong

          Hello, you are now looking at a book worm who likes animations about stickman, and is very nearly bad at math.
          If you wanna see my animations, search "AshFang1130" on BilliBilli(哔哩哔哩)
          P.S. I LOVE WARRIOR CATS!!!!!!!!
          P.P.S I also play GachaClub

          1 Reply Last reply Reply Quote 2
          • quacker88Q
            quacker88 MOD
            last edited by

            Hey @eeveelution ! Sorry for replying so late, but here's how I would look at your two problems:

            For the second one, since the problem only cares about the last digit of that huge sum, we only need to look at the last digit of each individual term!

            \(1^{15}\) is easy, the last digit of \(1^{15}\) is just \(1\).
            For the rest though, we don't want to expand it out (that's a lot of work).

            So, for \(2^{15}\), let's rewrite it as \((2^5)^3\).
            \(2^5=32 \implies \text{last digit is 2}\)
            \(2^3=8\)
            So the last digit of \(2^{15}=(2^5)^3\) is \(8\).

            Another way to look at it is to try and find a pattern among powers of 2. Let's do a few and see if we can find anything.
            \(2^1 \implies \text{last digit 2}\)
            \(2^2 \implies \text{last digit 4}\)
            \(2^3 \implies \text{last digit 8}\)
            \(2^4 \implies \text{last digit 6}\)
            \(2^5 \implies \text{last digit 2}\)
            \(2^6 \implies \text{last digit 4}\)
            Wait a minute, this is just gonna repeat \(2,4,8,6,2,4,8,6...\) It cycles every \(4\)!
            So, since \(15\) is \(3\) more than a multiple of \(4\), the last digit of \(2^{15}\) is going to be the \(3^{\text{rd}}\) number in the list, which is \(8\).

            You can repeat this process for the rest of the numbers.
            But, keep in mind that \(12\) has units digit \(2\), so \(12^{15}\) is going to have the same units digit as \(2^{15}\)!

            Also, one last tip-- since \(2012\) is so big, I would group everything in groups of \(10\). What I mean is, first find what the units digit of \(1^{15}+2^{15}+...+10^{15}\) is. Then, since \(11^{15}\) through \(20^{15}\) will have the same units digits (along with every other group of 10), you can use a shortcut and not have to calculate every single one.

            For the rectangle problem, I'm getting 79... which is weird... but either way I definitely think it's more than 73. What book is this?

            Desolate_101D 1 Reply Last reply Reply Quote 2
            • Desolate_101D
              Desolate_101 M0★ M1★ M2 M3 M4★ @quacker88
              last edited by

              @quacker88 thank you very much! I understand now, and also, for the second one it is a chinese book called 《动脑筋》.

              Hello, you are now looking at a book worm who likes animations about stickman, and is very nearly bad at math.
              If you wanna see my animations, search "AshFang1130" on BilliBilli(哔哩哔哩)
              P.S. I LOVE WARRIOR CATS!!!!!!!!
              P.P.S I also play GachaClub

              1 Reply Last reply Reply Quote 1

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