Download March 2010 Problem of the Month Solution

Okanagan College
Math Problem of the Month
MARCH 2010 – Solutions
MAIN PROBLEM: I have twelve envelopes, which are numbered 1 through 12, and twelve cards, which
are numbered 110 through 121. How can I place one card inside each envelope so that the number on each
envelope is a divisor of the number on the card inside it?
Solution: First list the factors between 1 and 12 of each of the numbers from 110 to 121:
110
1, 2, 5, 10, 11
116
1, 2, 4
111
1, 3
117
1, 3, 9
112
1, 2, 4, 7, 8
118
1, 2
113
1
119
1, 7
114
1, 2, 3, 6
120
1, 2, 3, 4, 5, 6, 8, 10, 12
115
1, 5
121
1, 11
The card with 113 written on it must be placed in envelope 1. But then the cards numbered 111, 115, 118,
119, and 121 must go in envelopes 3, 5, 2, 7, and 11, respectively. Then card 114 must go in envelope 6
because envelopes 1, 2, and 3 are full; card 116 must go in envelope 4; and card 117 must go in envelope 9.
The remaining envelopes are numbered 8, 10, and 12; and the remaining cards are 110, 112, and 120. The
only envelope left for card number 110 is number 10, and the only envelope left for card 112 is number 8,
leaving envelope 12 for card number 120. Then from this you can see that certain numbers can only have 1
possible outcome, so you end up by working backwards with the following:
# in
envelope
# on
envelope
1
6
5
4
9
2
7
12
11
113
114
115
116
117
118
119
120
121
BONUS PROBLEM: Two ferries start at the same time on opposite sides of a river. Then, they
travel across the water on routes perpendicular to the shores. Each ferry travels at a constant speed,
one faster than the other. They pass at a point 720 metres from the nearest shore. Both ferries remain
docked for 10 minutes before starting their return trip. On the way back, they meet 400 metres from
the other shore. How wide is the river?
Solution: 1760 metres. When the ferries meet for the first time, the combined distanced travelled by
the boats is equal to the width of the river. When they reach the opposite shore, the combined distance
is twice the width of the river. When they meet the second time, the total distance is three times the
river’s width. Since the boats move at a constant speed for the same length of time, then each boat has
gone three times as far as when they first met and had travelled a combined distance of one width of the
river. Since one ferry had travelled 720 metres when the first meeting occurred, its total distance at
the time of the second meeting must be 3 x 720 or 2160 metres. The second time, the distance is 400
metres more than the river’s width. Hence we subtract 400 from 2150 to obtain 1760 metres as the
width of the river.