Diagonalled.

Logic Level 3

1 2 3 4 5 6 7 □ 14 13 12 11 10 9 □ 8 15 16 17 18 19 □ 20 21 28 27 26 25 □ 24 23 22 29 30 31 □ 32 33 34 35 42 41 □ 40 39 38 37 36 43 □ 44 45 46 47 48 49 □ 56 55 54 53 52 51 50 □ 57 58 59 60 61 62 63 70 □ 69 68 67 66 65 64 71 72 □ 73 74 75 76 77 84 83 82 □ 81 80 79 78 ⋮ ⋮ ⋮ \large \begin{array}{c}1 & 2 & 3 & 4 & 5 & 6 & 7 & \square \\ 14 & 13 & 12 & 11 & 10 & 9 & \square & 8 \\ 15 & 16 & 17 & 18 & 19 & \square & 20 & 21 \\ 28 & 27 & 26 & 25 & \square & 24 & 23 & 22 \\ 29 & 30 & 31 & \square & 32 & 33 & 34 & 35 \\ 42 & 41 & \square & 40 & 39 & 38 & 37 & 36 \\ 43 & \square & 44 & 45 & 46 & 47 & 48 & 49 \\ \square & 56 & 55 & 54 & 53 & 52 & 51 & 50 \\ \square & 57 & 58 & 59 & 60 & 61 & 62 & 63 \\ 70 & \square & 69 & 68 & 67 & 66 & 65 & 64 \\ 71 & 72 & \square & 73 & 74 & 75 & 76 & 77 \\ 84 & 83 & 82 & \square & 81 & 80 & 79 & 78 \\ &\vdots &&&\vdots &&&\vdots \end{array}

Following the above pattern further down, which column will the number 15600 appear in?

Column 1 Column 2 Column 3 Column 4 Column 5 Column 6 Column 7 Column 8

Thành Đạt Lê by Thành Đạt Lê , 17, Singapore

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1 solution

Chew-Seong Cheong
Jul 12, 2018

We note from the table that the sequence has a period of 16 × 7 = 112 16 \times 7 = 112 . All we need to do is to find the remainder of any given number when divided by 112 and look at the table to find the column of where the remainder appears.

Now we have 15600 m o d 112 = 32 15600 \bmod 112 = 32 . And 32 appears in the Column 5 .

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