Page 2 of 3 FirstFirst 123 LastLast
Results 16 to 30 of 32

Math Help - Australian Maths Competition

  1. #16
    Newbie
    Joined
    Aug 2010
    Posts
    20
    Is this the answer?
    1053-764=289
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Quote Originally Posted by Username View Post
    What?? :O

    I don't get what you mean by that.
    Perhaps that the digits in the 4-digit number add up to form a number divisible by 9.

    For example, if you get 1089, you know it's divisible by 9 because 1+0+8+9 = 18 and 18 is divisible by 9.

    Since you have 1 and 0, the last numbers can be 8 and 9, or 6 and 2, or 5 and 3

    Sut since 2 was already used you are left with 8 and 9 or 5 and 3
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Username View Post
    Is this the answer?
    1053-764=289
    not quite!
    as 1053-784=269
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Newbie
    Joined
    Aug 2010
    Posts
    20
    Aww, so close! ha ha

    I assume 269 is the answer then?
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Quote Originally Posted by Archie Meade View Post
    not quite!
    as 1053-784=269
    So, is that it? It's the smallest possible result? Is there a way to make sure that it's the smallest value?
    Follow Math Help Forum on Facebook and Google+

  6. #21
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Unknown008 View Post
    So, is that it? It's the smallest possible result? Is there a way to make sure that it's the smallest value?
    1053-789=264

    That's the smallest I can find, so far.

    By reasoning, we can say the first 2 digits of the 4-digit number are 10
    and the first digit of the 3 digit number is 7.
    The first digit of the result is 2.

    20X, 21X are not solutions, neither is 22X.
    23X falls short because we must take 9 from 12 or 8 from 11.
    This could involve 8+carry from 12, 7+carry from 11.
    The remaining digits in the units positions will not combine.
    You can reason that way for 24X and 25X also, ending up at 26X which pulls through.

    Yikes! I missed the 24X..... here you go, eumyang
    Last edited by Archie Meade; August 7th 2010 at 07:06 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    Quote Originally Posted by Archie Meade View Post
    1053-789=264

    That's the smallest I can find.
    Actually, there is a smaller one:
    1\,0\,3\,5\:-\:7\,8\,9\:=\:2\,4\,6
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Newbie
    Joined
    Aug 2010
    Posts
    20
    246...
    If the answer can't be 20X to 23X, 246 is quite likely to be the lowest?
    Follow Math Help Forum on Facebook and Google+

  9. #24
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Username View Post
    246...
    If the answer can't be 20X to 23X, 246 is quite likely to be the lowest?
    Obviously not viable during competition, but easy brute force gives 246.

    Java:

    Code:
    public class DistinctDigSubtrQ {
        public static void main(String[] args) {
            int i,j;
            long t=time();
            String s;
            for(i=201;i<299;i++)
                if(hasDistinctDigs(s=str(i)))
                    for(j=1034-i;j<988;j++)
                        if(hasDistinctDigs(str(i+j)+str(j)+s)) {
                            System.out.println((i+j)+"-"+j+"="+i);
                            System.out.println("Elapsed: "+(time()-t)/1000.0+" s");
                            return;
                        }
        }
    
        static boolean hasDistinctDigs(String s) {
            boolean[] digs=new boolean[10];
            int i,d;
            for(i=0;i<s.length();i++) {
                d=Integer.parseInt(s.substring(i,i+1));
                if(digs[d])
                    return false;
                else digs[d]=true;
            }
            return true;
        }
        
        static String str(int n) {
            return Integer.toString(n);
        }
        
        static long time() {
            return System.currentTimeMillis();
        }
    }
    Output:

    Code:
    1035-789=246
    Elapsed: 0.04 s
    Last edited by undefined; August 7th 2010 at 08:20 PM. Reason: very small optimisation
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Newbie
    Joined
    Aug 2010
    Posts
    11

    More questions coming

    Question 29
    I have a list of thirty numbers where the first number is 1, the last number is 30 and each of the other numbers is on more than the average of its two neighbours. what is the largest number in the list

    Question 30 (I couldn't be stuffed reading this, didn't have time anyway)
    There are so many towns on the island of Tetra, all connected by roads. Reach town has three roads leading to three other different towns: one red road, one yellow road and one blue road, where no two roads meet other than at towns. If you start from any town and travel along red and yellow roads alternately (RYRY...) you will get back to your starting town after having travelled over six different roads. In fact RYRYRY will always get you back to where you started. In the same way, going along yellow and blue roads alternately will always get you back to the starting point after travelling along six different roads (YBYBYB). On the other hand, going along red and blue roads alternately will always get you back to the starting point after travelling along four different roads (RBRB). How many towns are there on Tetra?
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Newbie
    Joined
    Aug 2010
    Posts
    20
    Err.

    I don't really get 29 or 30...
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    Quote Originally Posted by chessweicong View Post
    Question 29
    I have a list of thirty numbers where the first number is 1, the last number is 30 and each of the other numbers is on more than the average of its two neighbours. what is the largest number in the list
    I think I got it. I noticed that as I put numbers into the sequence the difference between successive terms must decrease by 2 in order for the criterion to hold (each successive number is 1 more than the average of its 2 neighbors).

    I also noticed that after I get to a high point the numbers repeat themselves backwards, going back to 1. Seeing that the last number of this sequence is 30, I guessed that 30 was also the 2nd number.
    1, 30, ...

    1 + 29 = 30, so the next number has to be 30 + 27 = 57.
    1, 30, 57, ...

    57 works, because 30 is one more than the average of 1 and 57. The next number is 57 + 25 = 82.
    1, 30, 57, 82, ...

    And 57 is one more than the average of 30 and 82. You continue the pattern (adding 23 to 82, then adding 21, then adding 19,...), and after you add 1, you start subtracting 1, then subtracting 3, then subtracting 5, etc.

    Here is the complete sequence:
    1, 30, 57, 82, 105,
    126, 145, 162, 177, 190,
    201, 210, 217, 222, 225,
    226, 225, 222, 217, 210,
    201, 190, 177, 162, 145,
    126, 105, 82, 57, 30

    So the largest number is 226. I used the OpenOffice.org Calc spreadsheet again to figure this out, of course.
    Last edited by eumyang; August 7th 2010 at 07:53 PM.
    Follow Math Help Forum on Facebook and Google+

  13. #28
    Newbie
    Joined
    Aug 2010
    Posts
    20
    eumyang, you're a genius!
    Follow Math Help Forum on Facebook and Google+

  14. #29
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    I got number 30 too. I started with drawing hexagons... then got completely lost when I came to the last statement about the RBRB... I was thinking about a football, where hexagons were arranged together with pentagons, but this didn't fit the problem since following this logic, we would get hexagons and squares.

    My number 29 was a different though.
    Follow Math Help Forum on Facebook and Google+

  15. #30
    Newbie
    Joined
    Aug 2010
    Posts
    20
    Quote Originally Posted by Unknown008 View Post
    I got number 30 too. I started with drawing hexagons... then got completely lost when I came to the last statement about the RBRB... I was thinking about a football, where hexagons were arranged together with pentagons, but this didn't fit the problem since following this logic, we would get hexagons and squares.
    I'm trying to do 30 but, it doesn't seem to stop? :O

    EDIT: I think it's 24?

    Hexagon in the middle BYBYBY
    Centre is connected to 3 hexagons and 3 squares.
    The outer hexagons make 3 squares which those 3 squares connect together and make the whole island as all towns have 3 roads.
    Last edited by Username; August 7th 2010 at 09:52 PM.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 3 FirstFirst 123 LastLast

Similar Math Help Forum Discussions

  1. Question from the ICAS Maths competition
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: August 20th 2010, 08:23 PM

Search Tags


/mathhelpforum @mathhelpforum