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  3. Are You Smarter Than A Sixth Grader (Taking Seventh Grade Math)?

Are You Smarter Than A Sixth Grader (Taking Seventh Grade Math)?

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  • B Offline
    B Offline
    Bassam Abdul Baki
    wrote on last edited by
    #1

    Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?

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    • B Bassam Abdul Baki

      Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?

      Web - BM - RSS - Math - LinkedIn

      9 Offline
      9 Offline
      9082365
      wrote on last edited by
      #2

      8

      B 1 Reply Last reply
      0
      • B Bassam Abdul Baki

        Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?

        Web - BM - RSS - Math - LinkedIn

        G Offline
        G Offline
        gardnerp
        wrote on last edited by
        #3

        I get 8 paths. You can move up to the point where the sum of 1-n = 28. This happens at n = 7. (1 + 2 + 3 + 4 + 5 + 6 + 7 = 28). So the question is how many ways can you divide numbers 1-7 into two groups of 14? I count 8 ways.

        B 1 Reply Last reply
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        • 9 9082365

          8

          B Offline
          B Offline
          Bassam Abdul Baki
          wrote on last edited by
          #4

          What 8?

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          • G gardnerp

            I get 8 paths. You can move up to the point where the sum of 1-n = 28. This happens at n = 7. (1 + 2 + 3 + 4 + 5 + 6 + 7 = 28). So the question is how many ways can you divide numbers 1-7 into two groups of 14? I count 8 ways.

            B Offline
            B Offline
            Bassam Abdul Baki
            wrote on last edited by
            #5

            What 8?

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            • B Bassam Abdul Baki

              What 8?

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              G Offline
              G Offline
              gardnerp
              wrote on last edited by
              #6

              Didn't think I had to spell that one out: 761 752 743 7412 653 6521 6431 5432 For each of those they add up to 14. The missing numbers also add to 14.

              B 1 Reply Last reply
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              • B Bassam Abdul Baki

                Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?

                Web - BM - RSS - Math - LinkedIn

                L Offline
                L Offline
                Lost User
                wrote on last edited by
                #7

                Is it 8*8! ?

                B 1 Reply Last reply
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                • G gardnerp

                  Didn't think I had to spell that one out: 761 752 743 7412 653 6521 6431 5432 For each of those they add up to 14. The missing numbers also add to 14.

                  B Offline
                  B Offline
                  Bassam Abdul Baki
                  wrote on last edited by
                  #8

                  But you can only start with one step.

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                  • L Lost User

                    Is it 8*8! ?

                    B Offline
                    B Offline
                    Bassam Abdul Baki
                    wrote on last edited by
                    #9

                    That's assuming you make a wrong turn at Albuquerque. :laugh:

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                    L 1 Reply Last reply
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                    • B Bassam Abdul Baki

                      That's assuming you make a wrong turn at Albuquerque. :laugh:

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                      L Offline
                      L Offline
                      Lost User
                      wrote on last edited by
                      #10

                      Oh never mind, reread the question.. edit: 8 then. And since you seem to want to have them listed: R1 R2 U3 R4 U5 U6 R7 R1 R2 U3 U4 R5 R6 U7 R1 U2 R3 R4 U5 R6 U7 R1 U2 U3 U4 U5 R6 R7 U1 R2 R3 R4 R5 U6 U7 U1 R2 U3 U4 R5 U6 R7 U1 U2 R3 R4 U5 U6 R7 U1 U2 R3 U4 R5 R6 U7 the numbers are superfluous of course, but they prevent taking a wrong turn at Albuquerque..

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                      • B Bassam Abdul Baki

                        But you can only start with one step.

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                        G Offline
                        G Offline
                        gardnerp
                        wrote on last edited by
                        #11

                        That's completely irrelevant. 761 => 1U, 2R, 3R, 4R, 5R, 6U, 7U. 752 => 1R, 2U, 3R, 4R, 5U, 6R, 7U. 743 => 1R, 2R, 3U, 4U, 5R, 6R, 7U. 7412 => 1U, 2U, 3R, 4U, 5R, 6U, 7U. .... Follow my previous post. Simply put the 1 first instead of the order I have. I have them in a more mathematical order that I used to split the numbers into the two groups. Order does not matter.

                        B 1 Reply Last reply
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                        • G gardnerp

                          That's completely irrelevant. 761 => 1U, 2R, 3R, 4R, 5R, 6U, 7U. 752 => 1R, 2U, 3R, 4R, 5U, 6R, 7U. 743 => 1R, 2R, 3U, 4U, 5R, 6R, 7U. 7412 => 1U, 2U, 3R, 4U, 5R, 6U, 7U. .... Follow my previous post. Simply put the 1 first instead of the order I have. I have them in a more mathematical order that I used to split the numbers into the two groups. Order does not matter.

                          B Offline
                          B Offline
                          Bassam Abdul Baki
                          wrote on last edited by
                          #12

                          My bad. I missed a pair. :doh: :-O

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                          • B Bassam Abdul Baki

                            What 8?

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                            C Offline
                            C Offline
                            Corporal Agarn
                            wrote on last edited by
                            #13

                            7 8 9

                            Mongo: Mongo only pawn... in game of life.

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                            • B Bassam Abdul Baki

                              Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?

                              Web - BM - RSS - Math - LinkedIn

                              V Offline
                              V Offline
                              VuduDog
                              wrote on last edited by
                              #14

                              I guess I missed something. I thought 'n' was a specific number to be repeated at each step so the answer would be only the factors of 14. Therefore only these 'n' steps would be allowed. 1,2,7, and 14 - 2R, 2U, etc... or 7R,7U ... So the number of paths would be 4

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                              • B Bassam Abdul Baki

                                Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?

                                Web - BM - RSS - Math - LinkedIn

                                B Offline
                                B Offline
                                BinaryReason
                                wrote on last edited by
                                #15

                                At each point you can move either right or up. The setup makes the step sizes irrelevant to the problem - they will always add to 28 (14 horizontal + 14 vertical), (1 + 2 + 3 + 4 + 5 + 6 + 7). That means we will reach (14,14) from (0,0) in exactly 7 steps. At each of the 7 steps you can either move right or up, so you have two choices at every step node. The answer then becomes: 2^7 = 128 possible paths. If you were to visualize it, you would simply draw a complete grid between (0,0) and (14,14)

                                "There are only 10 types of people in the world - those who know binary and those who don't."

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                                • B Bassam Abdul Baki

                                  My bad. I missed a pair. :doh: :-O

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                                  T Offline
                                  T Offline
                                  TedFalasco
                                  wrote on last edited by
                                  #16

                                  Ha ha, I did too. I was with you all the way. I was all, "Yeah, you tell 'im, Bassam!" Until you said that and I rechecked my answers. :sigh:

                                  B 1 Reply Last reply
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                                  • T TedFalasco

                                    Ha ha, I did too. I was with you all the way. I was all, "Yeah, you tell 'im, Bassam!" Until you said that and I rechecked my answers. :sigh:

                                    B Offline
                                    B Offline
                                    Bassam Abdul Baki
                                    wrote on last edited by
                                    #17

                                    To err is human. To umm is herman.

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                                    • B Bassam Abdul Baki

                                      Homework problem. Q: On the (x,y) coordinate system, you start at (0,0) and want to get to (14,14). At each step n, you move exactly n steps either right or up. So initially, you can only move 1 unit right or up, then 2, then 3, etc. How many paths are there to (14,14)? A: He got one on his own and he understood the symmetry and got 2. I told him my answer and why. Anyone wants to try?

                                      Web - BM - RSS - Math - LinkedIn

                                      S Offline
                                      S Offline
                                      scott mankowitz
                                      wrote on last edited by
                                      #18

                                      Shouldn't it be 16? If there are 8 distinct groups of 3-4 numbers that sum to 14, one path will start with the first move going up and the next will start with the first move going right. Incidentally, here is my code:

                                      function combine(a, min) {
                                      var fn = function(n, src, got, all) {
                                      if (n == 0) {
                                      if (got.length > 0) {
                                      all[all.length] = got;
                                      }
                                      return;
                                      }
                                      for (var j = 0; j < src.length; j++) {
                                      fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
                                      }
                                      return;
                                      }
                                      var all = [];
                                      for (var i = min; i < a.length; i++) {
                                      fn(i, a, [], all);
                                      }
                                      all.push(a);
                                      return all;
                                      }

                                      function arr_sum(arr) {
                                      var n = 0;
                                      for (var i = 0; i < arr.length; i++) {
                                      n += arr[i];
                                      }
                                      return n;
                                      }
                                      k=[1,2,3,4,5,6,7];
                                      combine(k, 1).filter(function(v, i) {
                                      return 14 == arr_sum(v)
                                      })

                                      B 1 Reply Last reply
                                      0
                                      • S scott mankowitz

                                        Shouldn't it be 16? If there are 8 distinct groups of 3-4 numbers that sum to 14, one path will start with the first move going up and the next will start with the first move going right. Incidentally, here is my code:

                                        function combine(a, min) {
                                        var fn = function(n, src, got, all) {
                                        if (n == 0) {
                                        if (got.length > 0) {
                                        all[all.length] = got;
                                        }
                                        return;
                                        }
                                        for (var j = 0; j < src.length; j++) {
                                        fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
                                        }
                                        return;
                                        }
                                        var all = [];
                                        for (var i = min; i < a.length; i++) {
                                        fn(i, a, [], all);
                                        }
                                        all.push(a);
                                        return all;
                                        }

                                        function arr_sum(arr) {
                                        var n = 0;
                                        for (var i = 0; i < arr.length; i++) {
                                        n += arr[i];
                                        }
                                        return n;
                                        }
                                        k=[1,2,3,4,5,6,7];
                                        combine(k, 1).filter(function(v, i) {
                                        return 14 == arr_sum(v)
                                        })

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                                        B Offline
                                        Bassam Abdul Baki
                                        wrote on last edited by
                                        #19

                                        No. You always start with 1 step.

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                                        • B Bassam Abdul Baki

                                          No. You always start with 1 step.

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                                          S Offline
                                          S Offline
                                          scott mankowitz
                                          wrote on last edited by
                                          #20

                                          I think there are 16 distinct paths, as below R1U2R3R4U5R6U7 U1R2U3U4R5U6R7 R1R2U3U4R5R6U7 U1U2R3R4U5U6R7 R1R2U3R4U5U6R7 U1U2R3U4R5R6U7 U1U2R3U4R5R6U7 R1R2U3R4U5U6R7 U1U2R3R4U5U6R7 R1R2U3U4R5R6U7 U1R2U3U4R5U6R7 R1U2R3R4U5R6U7 R1U2U3U4U5R6R7 U1R2R3R4R5U6U7

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