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Australian Maths Competition [Calculators not Allowed]

  1. Apr 20, 2008 #1
    1. The problem statement, all variables and given/known data

    ok, i've got several questions:

    1. which integer is closest to [tex]\sqrt{1994 + \sqrt{1994}}[/tex]

    A. 44
    B. 45
    C. 46
    D. 47
    E. 48

    2. Four singers take part in a musical round of 4 equal lines, each finishing after singing the line through three times. The second singer begins when the first singer begins the second line, the third singer begins when the first singer begins the third line, the fourth singer begins when the first singer begins the fourth line. The fraction of the total singing time that all four are singing at the same time is:

    A. 3/4
    B. 3/5
    C. 2/3
    D. 5/6
    E. 8/15

    3. Our local baker sell small buns for 30c each, or 7 for $1, or $1.80 per dozen. My mother gives me $10 to bring home 60 buns and tells me I can keep the change. I want to be able to bring home at least 60 buns and have as much money left over as possible. What is the most change I can keep?

    A. $0.90
    B. $1.00
    C. $1.10
    D. $1.20
    E. $1.30

    4. Which pf the five numbers is not equal to any of the others?

    A. [tex]\frac{1996}{1997}[/tex]
    B. [tex]\frac{996}{997}[/tex]
    C. [tex]\frac{1997996}{1998997}[/tex]
    D. [tex]\frac{19971996}{19981997}[/tex]
    E. [tex]\frac{996996}{997997}[/tex]

    5. The number 119 is very curious.

    When divided by 2, it leaves a remainder of 1
    When divided by 3, it leaves a remainder of 2
    When divided by 4, it leaves a remainder of 3
    When divided by 5, it leaves a remainder of 4
    When divided by 6, it leaves a remainder of 5

    How many other 3-digit numbers have this property?

    A. 0
    B. 1
    C. 3
    D. 7
    E. 14

    2. Relevant equations



    3. The attempt at a solution

    1. i tried working with powers, i.e 1994[tex]^{\frac{1}{2}}[/tex] and then simplifying. i couldn't reach anything.

    2. I found the total time as 7 lines. (4 lines for the first singer and then 3 lines after him for the last singer). The only time all the four all singing together is when the first singer sings the last line.

    my answer [tex]1/7[/tex] is not ofund in the choices.

    3. I calculated the maximum change as $1.00 when he either buys 63 buns (7 for $1) or when he buys 60 buns ($1.80 the dozen) But i think it was a bit tooo easy.

    4. I don't know how to proceed

    5. Again, I dunno


    What is more important to me is learning the methodology to tackle these problems, so I need great help. thnks
     
  2. jcsd
  3. Apr 20, 2008 #2

    Gib Z

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    1) Well, [itex]\sqrt{1994} [/itex] comes up, and I tried approximating that first. A very rough approximation will do, because by squaring 44 and 45, you can see [itex]\sqrt{1994}[/itex] is somewhere in between there. So the expression in question is obviously at most [itex]\sqrt{1994+45} = \sqrt{2039}[/itex]. And we just squared 45 just a second ago, what do you think =D

    2) I'm a bit confused >.<" Do they all finish after singing 4 lines? What do you mean by, singing the line through 3 times? =S

    3) Ok Well remember, we can buy more than 60 if we need to save, eat the rest =] We can see the the 7 for $1 deal is the most economical, so try to get as many of those with barely going over. For $9, we get 63, more than we want. $1 saving isn't that much though.

    So try 8 bunches of the 7 deal, we get 56 for $8, now if we fill the rest with the dozen deal, we only get 20 cents. If we get the rest for the singles deal, we only have 80 cents, worse than before.

    So now try 7 bunches of the 7 deal. Then fill the rest with combinations of dozen and single deals, going just a bit over if you have to. I've only been doing it a few minutes and can already get us $1.20, see if you can get to that $1.30.

    4) The easiest way for this to realize that If any of these numbers are equal, then their difference from 1 will also be equal.

    So just subtracting, rewrite the options as:

    a) [itex]\frac{1}{1997}[/itex].

    b) [itex]\frac{1}{997}[/itex]

    c) [itex] \frac{1001}{1998997}[/itex]

    d) [itex]\frac{10001}{19981997}[/itex]

    e) [itex]\frac{1001}{997997}[/itex].

    Now if the numbers are to be equal, and their numerators are already equal, then their denominators should also be equal. And vice versa. Since there are 2 1001'ed numerators, and we can easily change a) and b) the be the same, multiply them by 1001/1001. You should be straight from here.

    5) OK Well we can rule out a whole huge bunch of cases quickly here! The fact that when its divided by 5, it gives remainder of 4 means that the last digit is either a 4 or a 9, because numbers divisible by 5 always end in either 0 or 5. But it can't end in 4, because then it would be divisible by 2, but its meant to be a remainder of 1! So now you only have to check 9 cases =]
     
  4. Apr 20, 2008 #3
    2) I interpreted singing four lines three times as having a total of 12 lines that the singers continuously cycle through. Remember that and you should be able to find the answer using your previous method.

    5) Consider the LCM of 2, 3, 4, 5, and 6. What is different about the LCM from the number you want?
     
  5. Apr 20, 2008 #4

    Hurkyl

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    You really should specify that the competition isn't happening right now. :grumpy:
     
  6. Apr 20, 2008 #5
    They don't allow calculators, but you can use the internet?
     
  7. Apr 20, 2008 #6

    cristo

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    I imagine the OP put "calculators not allowed" so that no-one replies and says "well just type the question in a calculator and see what the answer is." These are most probably practice questions, that he is looking for assistance with so as to prepare for the real thing.
     
  8. Apr 21, 2008 #7

    Gib Z

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    It's a real past paper from 1994, I vaguely remember doing it a few years ago as well. Many Pre Calc math tests in Australia do not allow calculators.
     
  9. Apr 21, 2008 #8
    bah yeah, the competition is for next month, i was just practicing some questions. Thank you Gib_Z for the method of proceeding.

    for number 2, what i understood was that the singer sings 3 times a line and there are four lines in all. the first singer obviously starts first, after he finishes the first line (all the 3 times i suppose) he starts the second line, and just at that moment the second singer starts his/her first line. and so on.....


    btw Gib_Z do you know if i can get past exam AMC questions on the internet? i've been googling a lot, but i couldn't find any. prolly it is copyright...
     
  10. Apr 21, 2008 #9
    @Kushal,

    I think #2 is referring to singing the song once through, then beginning again at the beginning, like a round.

    As for past AMC questions, you may find some of them at http://www.artofproblemsolving.com/
     
  11. Apr 22, 2008 #10

    Gib Z

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    Sorry I can't find any websites with previous AMC questions either! I know you can buy books with some of them from the Australian Mathematics Trust, search that up in Google and check out the products section.
     
  12. Apr 22, 2008 #11
    ooo alritz....thnks
     
  13. Apr 22, 2008 #12

    djeitnstine

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    I think you approaching it a tad wrong, because the question says all of them sing their lines 3 times AND when one finishes the other starts. The way I did it was to make a chart using the given algorithm and count the answer comes to be 9/15 or 3/5. However, there is an easier way of doing this. You know that the first singer will sing 4 lines 3 times, thus 12 times. But the last singer starts singing at the beginning of #1's last line and therefore will end 3 lines after, making 15. so there are 3 + 3 times all four are not in unison. so (6-15)/15 = 9/15 or 3/5

    Make a diagram, it helps. (L1 = line one (1); 1 = singer one (1), singer two (2) etc)

    L1 | 1 | 1,2,3,4 |1,2,3,4 |2,3,4

    L2 | 1,2 | 1,2,3,4 |1,2,3,4 |3,4

    L3 | 1,2,3 | 1,2,3,4 |1,2,3,4 |4

    L4 | 1,2,3,4 | 1,2,3,4 |1,2,3,4


    As for question 3, It says at least 60...not EXACTLY 60 so 61, or 62 is okay. Realizing this one finds that 7 for $1.00 is the cheapest ($0.14 per bun) So just multiply it by a high number close to 60. There are 2 candidates, x8 and x7. Lets try the first (x8) that gets us reasonably close ... 56 buns but we have 4 buns to worry about and the only other option is $0.30 per bun which which is $1.20 too much! that is (8 + 1.2 = 9.2) The latter gives us $7 (x7 or 49 buns) with 11 remaining..just add one dozen buns ($1.80) and you get $8.80. This leaves $1.20 change with 61 buns!
     
    Last edited: Apr 22, 2008
  14. Apr 22, 2008 #13

    djeitnstine

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    Nice deduction!
     
  15. May 14, 2009 #14

    kya

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    5. Well, i think that the numbers should end with 9.... because if it is not divisible by 5 giving a remainder 4, it should end with either 4 or 9. since ending with 4 will be divisible by 2, so only posibility is ending with a 9. Now, when divided by 6, remainder must be 5, for it to be so, the last number of the product of a number with 6 should be 4. on adding 4 with the 5, will get 9.
    example: only when last number is 9 x 6 , will give a number ending with 4, on adding the 5 with it, it will end with number 9,
    9x6=54. 54+5 give 59. and 59 has that property but it is now of 3 digit number,
    19x6=114, 114+5 gives 119. that number is already given above so exclude it from answer
    29x6=174, 174+5 gives 179.... and 179 has this property and is of 3 digit number,
    39x6=.......
    49
    59
    69
    79
    89
    99
    109
    119
    129............................................ proceed the same way,until exceeding 3 digit number as the final answer of the process. the final answer is 14, obtained by counting the numbers from 29, 39, 49 until the last one ^^"
     
  16. May 14, 2009 #15
    Kya, why on earth would you resurrect a year-old thread???? :rolleyes:


    01
     
  17. May 15, 2009 #16

    kya

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    lol, am new to PF, and through that question that i joined in ^^" can u help me to find similar questions... n how to do so??? :S
     
  18. Feb 3, 2011 #17
    1)Well I know this squaring technique:

    55 squared is like this:

    The last 2 numbers will always be 25 in squaring 2 digit numbers that end in five

    then your take the first number which is 5

    then add 1

    6 times the original number (5) = 30

    answer is 3025

    So that is nowhere near 1994
    so I try 45 which is 2025
    whic is very close, so my estimation of 1994 square rooted is about 44

    44+1994= 2038

    Now I have to square root that. 2038 is very close to 45 squared (2025) so my answwer to number 1 is 45

    2) imagine n is the amout of time to sing a line then I drew this (which is line a time line thingy) m is when 0 is when they do not

    4th singer 0 0 0 n n n n n n n n n n n n
    3rd singer 0 0 n n n n n n n n n n n n 0
    2nd singer 0 n n n n n n n n n n n n 0 0
    1st singer n n n n n n n n n n n n 0 0 0

    Now how long does it take from start to finish?
    15n
    Now how long were all of them singing at the same time? 9n
    so 9n/15n= 9/15= 3/5

    3) my answer is actually $1.60 but its none of the answers so yeah...

    4) cant be stuffed but I think its a

    5) okay this is a hard one but here is what I did; got a list of prime numbers up to 1000 (cause a number that can't be divided by numbers from 1-6 is basically a prime number) (then if my answwer is 5 I might have missed 2 that was not one)

    here is something that stood out: when divided by 5 remainder of 4

    now numbers that do that either ends with a 4 or a 9 but because no 3 digit prime number is even it is only 9 here are the numbers:

    109
    179 199 229 239 269
    349 359 379 383
    389 409 419 439 449 479 499
    509 569 599 619
    659 709 719 739
    769 809 829 839 859
    919 929
    now I know how to check if a number is divisible by 3 so for example I take 109-2 (beacuse it said remainder of 2)
    107= 1+0+7=8 8 is not a multiple of 3 which I can now eliminate



    179
    359
    419 449 479
    509 569 599
    659 719
    809 839
    929

    now I know how to figure out how to check for divisibility by 6= it has to pass 2 and 3 divisibility

    so for example I take 359 - 5 (because thats the remainder) =354 (it is even divisibility for 2 check) (3+5+4= 12 divisibility for 3 check) I will not cross this one out.

    I found out that all of them ahcieves this.

    I also know how to check a numbers divisibility for 4= check the last 2 digits and see if they are divisible by for (I know my 4 times tables in the 2 digits so thats gud)

    ...... same as the others (cant be stuffed to type it up again)...

    I found out that they all work..

    Now the only one left is 2 but i already know this works on all beacuse all odd numbers can be divisible bytwo if you take 1 away. There is 13 numbers left I problbly made a mistake and deleted one that wasn't supoosed to be deleted. There fore my answer is

    e
     
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