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October 2006 Solution
1. There are four answers for Puzzler 1 and they are:
2,438,195,760 3,785,942,160
4,753,869,120 4,876,391,520
For a number to be divisible by each number 1 through 18, the number must be a multiple of the least common multiple of the numbers from 1 through 18.
The least common multiple of all numbers from 1 through 18 is
(2^4)*(3^2)*5*7*11*13*17 = 12,252,240.
Because we must obtain a 10-digit number (that includes each digit 0 – 9), we must multiply 12,252,240 by a number between 9 and 816.
After checking each possible multiplier between 9 and 816 we arrived at the four solutions given above.
Note: one could apply certain divisibility rules to reduce the number of cases to check.
2. Answer:
Bag one has 27 marbles; Bag two has 24 marbles; Bag three has 18 marbles; Bag four has 17 marbles; Bag five has 14 marbles.
To find the number of marbles in bag one we use the fact that bags two and three contain a total of 42 marbles while bags four and five contain a total of 31 marbles. Since there are 100 marbles total, bag one must contain 100 – (42+31) = 27 marbles.
Now bags one and two contain a total of 51 marbles and thus bag two must contain 51 – 27 = 24 marbles. Because bags two and three contain a total of 42 marbles, bag three must contain 42 – 24 = 18 marbles. Because bags three and four contain a total of 35 marbles, bag four must contain 35 – 18 = 17 marbles. Because bags four and five contain a total of 31 marbles, bag five must contain 31 – 17 = 14 marbles.
3. Answer:
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6 |
2 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
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A |
B |
C |
D |
E |
F |
G |
H |
I |
J |
The 6 in position A indicates that there are six 0’s in the number; the 2 in position B indicates that there are two 1’s; the 1 in position C indicates that there is one 2; the 1 in position G indicates that there is one 6; the 0 in positions D, E,F,H,I, and J indicate that there are no 3’s, 4’s, 5’s, 7’s, 8’s, and 9’s respectfully.
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