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4th grade math problem

  1. Oct 5, 2006 #1
    My daughter just came home with this problem and I feel really stupid cause I have looked it over for an hour and cannot get it. I just finished a calc II refresher course and maybe I am overthinking(nuking) the problem. Sure, I can randomly plug and chug some numbers through my calculator, but was wondering if someone would provide a little insight so I can explain to her how to do it?

    In this puzzle, three digits are provided as clues. Fill in the boxes(X) [DECIMALS ARE ONLY PLACE HOLDERS TO GET THE CORRECT FORMATTING] below with the prime numbers, 2,3,5,or 7 so the multiplication makes sense.

    Please forgive me for the poor formatting of the equation

    ..XXX
    ....XX
    EQUALS
    .XXXX
    2XXX.
    EQUALS
    2X5XX
     
    Last edited: Oct 5, 2006
  2. jcsd
  3. Oct 5, 2006 #2

    CRGreathouse

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    Of those digits, only 5 can appear as the units digit. That gives you two Xs right off the bat.
     
  4. Oct 5, 2006 #3
    I don't understand the comment "units" digit. PLease clarify if you have time
     
  5. Oct 5, 2006 #4

    Office_Shredder

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    If you have a number:

    abcd

    where a, b, c and d are the digits, then d is the units digit, c is the tens digit, b is the hundreds digit, etc.

    you just need a 5 in the units digit of each solution, and one of the two initial units
     
  6. Oct 5, 2006 #5

    why can 5 be the only units digit in each solution
     
  7. Oct 6, 2006 #6

    arildno

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    I misunderstood your notation, sorry.
    All deleted.
    I'll get back with a better reply.
     
  8. Oct 6, 2006 #7

    I am sorry for the poor formatting.

    There is to be only one multiplication involved. the subsequent X's below the first(equals) are the products of the multiplication with the 5 digit final result being the answer.
     
  9. Oct 6, 2006 #8

    arildno

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    Here's how you should start, gocnavy:
    Get a better notation than the one in the book first!!
    So I'll call the five numbers you are to find A,B,C,D and E.
    We have:
    [tex]A=A_{h}*100+A_{t}*10+A_{u}*1[/tex]
    [tex]B=B_{t}*10+B_{u}*1[/tex]
    [tex]C=C_{th}*1000+C_{h}*100+C_{t}*10+C_{u}*1[/tex]
    [tex]D=2*1000+D_{h}*100+D_{t}*10+D_{u}*1[/tex]
    [tex]E=2*10000+E_{th}*1000+5*100+E_{t}*10+E_{u}*1[/tex]
    In addition,
    we have the relations:
    [tex]A*B_{u}=C[/tex]
    [tex]A*B_{t}=D[/tex]
    [tex]D*10+C=E[/tex]


    The indices th, h,t,u stands for "no. of thousands", "no. of hundreds", "no. of tens" and "no. of units", respectively,
    and the digits [tex]A_{h},A_{t}, A_{u},B_{t},B_{u},C_{th},C_{h},C_{t},C_{u},D_{h},D_{t},D_{u},E_{th}, E_{t},E_{u}[/tex]
    can only be 2,3,5 or 7

    Got that?
     
  10. Oct 6, 2006 #9

    arildno

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    I have seen your reply; what I've written in the previous post is concerned with how you meant the problem to be understood.
     
  11. Oct 6, 2006 #10

    arildno

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    Now, we have that [itex]E_{u}=C_{u}[/itex]
    and we must also have:
    [tex]E_{u}=C_{u}=D_{u}=5[/tex]
    The reason for this, is that the only PRODUCTS of 2,3,5,7 that has as its last digits 2,3,5 or 7 is 3*5, 5*3, 5*5 and 7*5, all of whom having last digit 5.

    Thus, we have the schema (in your notation):
    ..XXX A
    ....XX B
    -----
    .XXX5 C
    2XX5. D
    ------
    2X5X5 E
    Looking at the next last column, we see that we must have:
    a) [tex](C_{t},E_{t})=(2,7)[/tex]
    or:
    b)[tex](C_{t},E_{t})=(7,2)[/tex]
     
    Last edited: Oct 6, 2006
  12. Oct 6, 2006 #11
    I understand this. Thank you very much. BUT is my 4th grade daughter expected to follow this same format? Or do you see this as a plug and chug method of a math game to stumble upon the correct answer? This seems far to complicated to expect a 4th grader to do. Is there possibly a more elementary approach to the problem? I sent her teacher a npte, and her reply was the 3digit number begins with a 7 and the two digit number starts with 3.
     
  13. Oct 6, 2006 #12

    arildno

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    I am horrified that this type of exercise is given to a 4th grader (I'm not sure, but that's age group 10-11, right?).

    If the kid is to do this METHODICALLY, and systematically, then this is way too cumbersome for the average kid. That is the way I wanted to show you, but I wanted some response before proceeding.

    I assume this is a meant as a guess-and-play exercise, that is something wholly without pedagogical value that only will be fun for those few A+ kids who like to play around with numbers.

    In my opinion, kids ought to master prescribed algorithms and develop understanding of what they are doing, they are simply not competent enough in math to find "clever shortcuts" or new ways to do math.
     
  14. Oct 6, 2006 #13

    arildno

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    Okay, now that I've cleared out a few points, we must now look in detail on the multiplication of A with [itex]B_{t}[/itex] (producing C) and [itex]B_{u}[/itex] (producing D), respectively. Since the arguments will walk along the same lines for both numbers, I'll work with a "b" that can be 2,3,5 or 7.

    We therefore have:
    [tex]A*b=A_{h}*b*100+A_{t}*b*10+A_{u}*b (EQ 2)[/tex]
    This must be a 4-digit number ending as I've said in 5, and thus b=3,5 or 7.
    We will treat the cases b=7 and b=3 first, since these forces [itex]A_{u}=5[/itex]

    b=7:
    We insert b=7 in (EQ 2) and rearrange it:
    [tex]C=A*7=7A_{h}*100+(7A_{t}+3)*10+5*1[/tex]
    Thus, if b=7, then we must have [itex]A_{t}=2 or 7[/itex] by looking at the 10's place.
    b=7,At=7:
    We now have:
    [tex]C=(7A_{h}+5)*100+2*10+5*1[/tex]
    Note that for NO choice for Ah will we get a proper 100-digit for C!
    Thus, if b=7, then At cannot be 7, but must be 2!
    b=7, At=2:
    We get, by inserting these in (EQ 2) and rearranging:
    [tex]C=(7A_{h}+1)*100+7*10+5*1[/tex]
    Here, we see that only values 2 and 3 are valid for Ah.

    Thus, consistent with b=7, we have the possibilities:
    A=225, C=1575 and A=325, C=2275
    (Note that if [itex]B_{t}=7[/itex], then A MUST be 325, since the first digit in D is 2!)

    Now, we may continue..(sigh)
     
  15. Oct 6, 2006 #14

    arildno

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    Now, note that if B=77 (and therefore A=325),
    then:
    .2275
    2275
    ------
    25025,
    that is, not both the digits of B can be 7.
    Let us therefore look at the two other cases with Bt=7, namely B=73 and 75, and A=325
    B=73 can be dismissed outright, since 3*325=975, which cannot work for C.
    B=75 doesn't work either, for that yields C=1625

    Thus we have shown that Bt must either be 3 or 5, it cannot be 7!
     
  16. Oct 6, 2006 #15
    wow...my daughter is 9 and still has some difficluty doing long division. She does not enjoy math enough yet for a problem like this. Her attention span with respect to mathematics is very minimal.
     
  17. Oct 6, 2006 #16

    arildno

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    Quite as I thought, a perfectly normal 9 year old.

    As you can see, I am systematically going through the options, there really isn't anything to learn from such exercises, they are just tests for your endurance of numbers.
     
  18. Oct 6, 2006 #17

    arildno

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    As an aside, she really ought to be given time by the teachers to work on those long divisions, that is important stuff, whereas the given exercise is both tedious and unimportant.
     
  19. Oct 6, 2006 #18
    I agree. She is in Venture(gifted) classes on Wednesdays and is out of her normal class room. Her primary teacher gave her this on Wednesday as home work because she missed class. She is a brilliant kid, but her gift is more for the arts. She taught herself to read by the age of three. Mathematics does not come as easy for her. Thank you for your time and effort to explain the procedure for tackling such a problem. I can see where it may draw on the interest of a child who enjoyed playing with numbers. Do we have a final answer yet?
     
  20. Oct 7, 2006 #19

    arildno

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    Give me some time. It is rather boring to do such an exercise and type it in, so I must take a break from time to time.
     
  21. Oct 7, 2006 #20
    I understand and did not intend to rush you.
     
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