 Puzzle thread Nov. 24 |
Mon, 24 November 2003 21:04  |
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 Ron |
 | Commander
Forum Administrator
Stars! AutoHost Administrator |
Messages: 1231
Registered: October 2002
Location: Collegedale, TN | |
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Here are this week's puzzles. 24 hour deadline to Private Message me your answers.
1. Examine carefully the following sequences of numbers:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
Although the sequences appear to behave totally at random, in fact, after the first number, each sequence is constructed in a precise and methodical way based on the previous one.
What is the next sequence?
2. A whole number is said to be square if it is equal to some whole number multiplied by itself. The following are square numbers: 25 (5x5) and 144 (12x12).
How must one place the numbers 1 to 15 on the blank lines below (without repeating any of them) in such a way that the sum of the numbers in any two consecutive spaces is always a square?
__ __ __ __ __ __ __ __ __ __ __ __ __ __ __
3. A certain whole number, whose last digit is 7, has the curious property that in order to multiply it by 7, all that is needed is to take the 7 from its right end and place it at the beginning.
What is the number?
4. A teacher writes a whole number less than 50000 on the chalkboard. One student states that the number is a multiple of 2; a second student states that the number on the board is a multiple of 3; and so on until a twelfth student says that the number on the board is a multiple of 13. The teacher remarks that all except 2 of the students were right, and morever, that the two who were wrong spoke one right after the other.
What was the number that the teacher wrote on the chalkboard?
5. X is a 3-digit number. If you reverse the order of the digits, and insert a digit, the new number is triple the old number.
What is X?
Thread locked until dealine expires.
Ron Miller
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| Re: Puzzle thread Nov. 24 |
Tue, 25 November 2003 22:13  |
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| Ron |
| Commander
Forum Administrator
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Messages: 1231
Registered: October 2002
Location: Collegedale, TN | |
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Wow! That's what I get for making this week's puzzles too easy... 6 people submitted answers, and ALL 6 got them all right! 
The people who got all the puzzles correct are: BlueTurbit, FurFuznel, overworked, Ashlyn, LEit, and Mazda.
And the random number program picks Mazda as this weeks Puzzle Master. 
Here are answers to the puzzles:
1. Starting with the second sequence, each sequence is simply a detailed description of the previous one. So, for example, the second sequence describes the first one: one one; that is, 1, 1. The third sequence describes the second: two ones; or 2,1. The forth describes the third: one two, one one; or 1,2,1,1. etc.
So the answer is 11131221133112132113212221
2. Two possible answers for this one:
9 7 2 14 11 5 4 12 13 3 6 10 15 1 8 and reversed.
3. The answer is: 1014492753623188405797
4. The number the teacher wrote on the chalkboard was 25740.
Let N be the number we are looking for. Since all the students except two who spoke one after another, were correct, it can be deduced that N can be divided by 1,2,3,4,5,6,10,11,12, and 13. This is because if 2 does not divide N, neither does 4; if 3 does not divide N, neither does 6; if 5 does not divide N, neither does 10; and so on. All of this leaves 7,8, and 9 as the only possible numbers that do not divide N. It is therefore necessary to examine two cases:
Case 1. N is not divisible by 8 and 9. In this case, however, the smallest number divisible by all the other numbers up to 13 (ie, their least common multiple) is 60060, a number greater than 50000.
Case 2. N is not divisible by 7 and 8. The least common multiple of the remaining numbers is 25740, and since any other multiple of the same numbers is greater than 50000, this must have been the number written by the teacher.
5. The 3-digit number is 351.
Thread unlocked.
...
Ron Miller
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| Re: Puzzle thread Nov. 24 |
Wed, 26 November 2003 06:01 |
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| mazda |
 | Lieutenant |
Messages: 655
Registered: April 2003
Location: Reading, UK | |
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* mazda gently bows *
I think that Ron's random number generator deserves most of the credit.
Also, I have another problem if people would like to have a go.
The "equation" N x 86 = 6N8
(i.e. swap the 86 around and place the original number in the middle)
has a trivial solution in N = 8
8 x 86 = 688
What is the next integer solution for N ?
(are these called Diophantine equations or something ?)
[Updated on: Wed, 26 November 2003 08:11] Report message to a moderator
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| Re: Puzzle thread Nov. 24 |
Thu, 27 November 2003 03:23 |
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| mazda |
| Lieutenant |
Messages: 655
Registered: April 2003
Location: Reading, UK | |
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| gible wrote on Thu, 27 November 2003 02:17 |
I believe diophantine equations are polynomials with positive integer solutions...
so theoretically you could turn it into a diophantine equation by expressing it as 86n=608+10n
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Thanks.
BTW, you've just posted (I believe) the easiest method of solving it !
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