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Math Help - Further Problem Solving

  1. #1
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    Further Problem Solving

    Hello again MHF
    here's another question that I am struggling with:
    "We say a number is ascending if its digits are strictly increasing. For exmaple,189 and 3468 are ascending while 142 and 466 are not. For which ascending 3-digit number n (between 100 and 999) is 6n also ascending?"

    ALSO if you could have a look at:
    Code:
    http://docs.google.com/viewer?a=v&q=cache:VUkDh8z6kwQJ:www.chiuchang.org.tw/modules/mydownloads/visit.php%3Flid%3D286+we+say+a+number+is+ascending+if+its+digits+are+strictly+increasing.+For+exmaple,189+and+3468+are+ascending+while+142+and+466+are+not.+For+which+ascending+3-digit+number+n+%28between+100+and+999%29+is+6n+also+ascending%3F&hl=en&gl=au&pid=bl&srcid=ADGEEShEbG423mHPBDUpCzh7NCiRak0S4nF_qRQkltAkGjbrM7UVZBYr-5Qw8MazU4HLQCMYZEgsCkwkxH1_SxBmkVm73bXW7wtoLunzJiYYmsDUWjhrPeir8NbfHwwPMYpNNN6RbEeK&sig=AHIEtbR6FdfPv58Qtfh1pIemcOsvUPJKvg
    Question 23 is there any logical way to go about doing that, I having trouble folding it up and unfolding it in my head :O (please don't say draw map out and start doing it!), also you may want to have a look at the last question, Question 30
    That seems insane..
    Thanks again!
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  2. #2
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    Quote Originally Posted by 99.95 View Post
    Hello again MHF
    here's another question that I am struggling with:
    "We say a number is ascending if its digits are strictly increasing. For exmaple,189 and 3468 are ascending while 142 and 466 are not. For which ascending 3-digit number n (between 100 and 999) is 6n also ascending?"
    A quick analysis and you'll find out that it's best not to have (6 times digit > 10). So 123 is a logical choice.
    ALSO if you could have a look at:
    Code:
    http://docs.google.com/viewer?a=v&q=cache:VUkDh8z6kwQJ:www.chiuchang.org.tw/modules/mydownloads/visit.php%3Flid%3D286+we+say+a+number+is+ascending+if+its+digits+are+strictly+increasing.+For+exmaple,189+and+3468+are+ascending+while+142+and+466+are+not.+For+which+ascending+3-digit+number+n+%28between+100+and+999%29+is+6n+also+ascending%3F&hl=en&gl=au&pid=bl&srcid=ADGEEShEbG423mHPBDUpCzh7NCiRak0S4nF_qRQkltAkGjbrM7UVZBYr-5Qw8MazU4HLQCMYZEgsCkwkxH1_SxBmkVm73bXW7wtoLunzJiYYmsDUWjhrPeir8NbfHwwPMYpNNN6RbEeK&sig=AHIEtbR6FdfPv58Qtfh1pIemcOsvUPJKvg
    Question 23 is there any logical way to go about doing that, I having trouble folding it up and unfolding it in my head :O (please don't say draw map out and start doing it!), also you may want to have a look at the last question, Question 30
    That seems insane..
    Thanks again!
    23) If you have trouble visualising this, try 'mapping' each column and row to the letters. For example, looking at the second picture, Column 5 > Column 4 > Column 3 > Column 2 > Column 1. Do the same thing with the rows. Looking at the first picture, row 1 > row 2 > row 3 > row 4. Hint: Sort by rows first. If there are multiple entries in the same row, sort those by columns.

    30) Let the total number of rabbits initially be n and the number deposited at each house is x.

    After the first house, we have n-x rabbits. We cross the bridge to double the number we have. So we now have 2n - 2x rabbits

    After the second house, we have (2n - 2x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(2n - 3x) = 4n - 6x rabbits

    After the third house, we have (4n - 6x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(4n-7x) = 8n - 14x rabbits

    After the fourth house, we have (8n - 14x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(8n-15x) = 16n - 30x

    At the fifth house, we have (16n-30x) - x rabbits, which equals 0 as we now have no more rabbits.

    So 16n - 31x = 0. To find the minimum number of rabbits required, think about this logically:

    the least amount of rabbits he can leave at each house is 1. So let x = 1. If x = 1, then 16n = 31. n = 31/16. But this is not a whole number, so x cannot equal 1.

    Instead of checking every single integer up, we can look at why x = 1 doesn't work -> 31 is not divisible by 16. So if we look for the least common multiple of 16 and 31, we can solve for n easily.

    LCM of 31,16 = 496. Therefore we need 31x = 16n = 496 => (x = 16, n = 31)

    This can be extended to any number m houses. If you look at the pattern generated on each step, the equation for the amount of rabbits is:

     2mn - (2^m - 2)x - x = 0

    Where m = number of houses; n = initial number of rabbits; x = rabbits deposited at each house.
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  3. #3
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    Quote Originally Posted by 99.95 View Post
    Hello again MHF
    here's another question that I am struggling with:
    "We say a number is ascending if its digits are strictly increasing. For exmaple,189 and 3468 are ascending while 142 and 466 are not. For which ascending 3-digit number n (between 100 and 999) is 6n also ascending?"
    I think you'll have to use a certain amount of trial and error here, but you can cut down on the number of possibilities.

    Say that the original number n has digits xyz, where x<y<z. The number 6n could have three or four digits, so call its digits abcd, where a=0 if it is a three-digit number, and a<b<c<d. Then z must be at least 3, and in fact the only way that z could be equal to 3 would be if n = 123. But then 6n = 738, which is not an ascending number. Therefore z must be at least 4. The same applies to d, which is also the final digit of a multiple of 6, so it must be 4, 6 or 8.

    So the smallest possible candidate for n is n = 124. But then 6n = 744, which has a first digit of 7. The only ascending 3-digit number starting with 7 is 789, which is not a multiple of 6, so that tells us that 6n must be a 4-digit number. The only ascending 4-digit number ending in 4 is 1234, which is not a multiple of 6. Therefore d cannot be 4, and we must have d = 6 or 8.

    The number 6n = abcd is thus an ascending 4-digit number ending in a 6 or an 8. It is a multiple of 6, and so its digits must add up to a multiple of 3. That cuts down substantially on the number of possibilities, and I found at that stage that it was easiest just to write down a list of all such numbers. They are
    1236
    1356
    2346
    3456
    1248
    1278
    1368
    1458
    1578
    2358
    2478
    2568
    3468
    4578

    Divide each of these by 6 and check whether the answer is an ascending number. Only one of them is!

    Spoiler:
    3468 = 6\times 578
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  4. #4
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    Quote Originally Posted by Gusbob View Post
    30) Let the total number of rabbits initially be n and the number deposited at each house is x.
    After the first house, we have n-x rabbits. We cross the bridge to double the number we have. So we now have 2n - 2x rabbits
    After the second house, we have (2n - 2x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(2n - 3x) = 4n - 6x rabbits
    After the third house, we have (4n - 6x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(4n-7x) = 8n - 14x rabbits
    After the fourth house, we have (8n - 14x) - x rabbits. We cross the bridge to double the number we have. So we now have 2(8n-15x) = 16n - 30x
    At the fifth house, we have (16n-30x) - x rabbits, which equals 0 as we now have no more rabbits.

    So 16n - 31x = 0. To find the minimum number of rabbits required, think about this logically:

    the least amount of rabbits he can leave at each house is 1. So let x = 1. If x = 1, then 16n = 31. n = 31/16. But this is not a whole number, so x cannot equal 1.

    Instead of checking every single integer up, we can look at why x = 1 doesn't work -> 31 is not divisible by 16. So if we look for the least common multiple of 16 and 31, we can solve for n easily.

    LCM of 31,16 = 496. Therefore we need 31x = 16n = 496 => (x = 16, n = 31)

    This can be extended to any number m houses. If you look at the pattern generated on each step, the equation for the amount of rabbits is:

     2mn - (2^m - 2)x - x = 0

    Where m = number of houses; n = initial number of rabbits; x = rabbits deposited at each house.
    Thanks for the reply both of you, I think for now i'll focus on the rabbits and get back to ascending digits later.
    So what I've understood what you're telling me for the most of it however the equation for the last part isn't really necessary is it?
    The answer to this question is 31, which is the amount of x values you had, however why does that mean that the minimum amount of rabbits in the first house is 31. After I used your equation on the first house i obtained and answer of 62.
    Thanks

    Edit: also may I add - for ascending digits question - that this was from a competition paper and that obviously we are given a time limit so the faster we work the better, and as for the ascending question it looks quite time consuming, is there no short cut to this one?
    Thanks again
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