Bunuel wrote:

Starting at M, Bob counts each vertex as he moves a chip clockwise around the inside of the square shown above. Ann also starts at M, but moves the chip in a counter-clockwise direction around the outside of the square. They count consecutive vertices together in unison 1, 2, 3, 4, 5, 6, etc. , and their positions for the first four counts are indicated above. Where will their chips be when both count 15?

(A) Both at M

(B) Both at K

(C) Bob's at M and Ann's at K

(D) Bob's at L and Ann's at J

(E) Bob's at J and Ann's at L

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I think Answer B, both at K.

Easiest way: Count. Count clockwise, and then counterclockwise. No kidding. Just remember to count "1" at M.

Why do they both land on K at count 15? I can think of four approaches that do not involve trigonometry. Here are two. (Neither seems quick to lay out, but I have tried. Describing patterns can be hard.)

CyclicityOne approach is cyclicity. Bob and Ann have to land on a vertex. Take the letter from the vertex AND the "count" number from the first time around the square. Use the letter and the count value to list ordinal terms.

Bob's cycle

Letter = nth term = count value

M = \(x_1 = 1\)

J = \(x_2 = 2\)

K = \(x_3 = 3\)

L = \(x_4 = 4\)

M = \(x_5 = 1\)

J = \(x_6 = 2\)

The cycle repeats after 4, i.e., cyclicity is 4.

15/4 = 3 remainder 3. Like units digits' patterns, the 15th term (the 15th count), because it is R3, will have the same value as the third term: \(x_3 = 3 = K\).

This cyclicity is analogous but not identical to that of units digits.

The first time around, the terms have corresponding cardinal and ordinal values: 1 = 1st term, 2 = 2nd term, etc. After that first time around, cardinal values repeat, ordinal values do not, but the ordinal values follow the cardinal numbers' pattern.

Bob and Ann will meet at the 1st, 5th, 9th, and 13th terms. They will also meet at the 3rd, 7th, 11th, and 15th terms. Because they are going different directions, they will not meet on the even numbered terms.

Ann's cycle:

M = \(x_1 = 1\)

L = \(x_2 = 2\)

K = \(x_3 = 3\)

J = \(x_4 = 4\)

M = \(x_5 = 1\)

Identical to Bob's on the cyclicity front. On the 15th count, because 15/4 leaves remainder 3, both land on the vertex of the third term in the cycle. That is K.

Intervals and tick marks Imagine the vertices as tick marks on a number line, and the sides as intervals.

There is one more tick mark than there are intervals. Here is one time around the perimeter:

Square: |---|---|---|---|

Labels: M---K---J---L---M

The number of

intervals from M to M is four. But the number of tick marks from M to M is five. There is

one fewer interval than there are tick marks. The interval is the length of a side. The tick?

Each time Bob and Ann hit a vertex, they count. That count, at each vertex, is a tick mark. 15 tick marks will have 14 intervals.

So a count of 15 tick marks means they cover the distance of 14 intervals.

To figure out where covering 14 intervals ends, take the total number of intervals and divide by the number of intervals per cycle around the square.

From the figure above: one cycle around the square is 4 intervals (whose length = square's side).

\(\frac{14}{4}\) = 3 cycles plus remainder 2 intervals.

There are three full cycles of M to M. There are two side lengths left over.

With a square figure of four sides, two side lengths away from an original vertex -- irrespective of direction traveled -- lands you on the diagonally opposite vertex. From M, where they both start, that diagonally opposite vertex is K.

Answer B

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At the still point, there the dance is. -- T.S. Eliot

Formerly genxer123