Real First Grade Math Problem (which I can't answer)?

One of my kids brought home this problem. Ref: http://i278.photobucket.com/albums/kk114/Remo_Avir...

First, this is really tough in principle. I'm surprised that it is from first grade. Second, I can't seem to get it correct. I think it is impossible -- or am I missing something?

Update:

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thanks nino,

I too came to the conclusion that it was unsolvable.

Update 2:

and thank you Baird, too.

Update 3:

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I saw my kid's teacher today and said there was no answer to the problem. (I had picked the problem up at an open house as an example of the work the kids do in class).

He said, "I know, I've e-mailed the company a few times and they even put out a new edition of the book with the same mistake"

He went on to say that "the kids work on these problems as a group in class, and when they came to this particular problem they concluded that there was no answer"

Pretty interesting stuff for first grade.

**********

Thank you everyone. I thought I'd made a mistake when I concluded it could not be solved.

I'll pick best answer tomorrow (unless I'm chasing kids all day which often happens) or Sunday at the latests

14 Answers

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  • 1 decade ago
    Best Answer

    I think I have a simple proof of its impossibility.

    E = even, O = odd

    There are 3 possible types of equation:

    E ± E = E

    E ± O = O

    O ± O = E

    Note there is no equation containing exactly 2 Es.

    We are given 4 Es and 5 Os. How do we place the 4 Es? I claim it's impossible to put 4 Es in the pattern without requiring that a 5th E be added. That is,

    1. If you put 3 Es such that they all are in different rows and columns (e.g., on a diagonal), adding the 4th will result in a row and a column, each with 2 Es and thus requiring another, making at least 6.

    2. If you put 2 Es in the same row or column (r/c), you must put the 3rd E in that r/c. Adding the 4th will produce another r/c with 2 Es, requiring a 5th.

    3. And obviously if you put 3 Es in the same r/c, adding the 4th will also require a 5th.

    If sfguybryan shows his solution with two 6s he'll blow my argument out of the water.

  • 1 decade ago

    The easiest way to prove the impossibility of an answer is to consider all possible ways odd and even numbers can add or subtract in a 3x3 array. We can examine all possible combinations of 0s and 1s in one 2x2 array (16 in all), in which the remaining 5 squares are completely determined by the 0s and 1s in the 2x2 array. Regardless of where the addition, subtraction, and equal signs are, there cannot be exactly 5 odd numbers in this 3x3 array. (There only can be 0, 2, 4, or 6 odd numbers). Hence, since that's how many odd numbers there are given in the problem, no solution exists, no matter how the "math" is rearranged in the 3x3 array.

    Since it's impossible for the given set {2,3,3,5,6,8,8,9,11}, even with other operation combinations, the question now becomes, "what's the simplest possible mistake to have resulted in an impossible problem?' After some diddling around, the simplest is that the 5 should have been a 6. With the revised set {2,3,3,6,6,8,8,9,11} (FOUR odd numbers), a solution exists as follows:

    3 + 8 = 11

    +....-.....-

    6 + 2 = 8

    =...=....=

    9 - 6 = 3

  • 1 decade ago

    in any row or column with - and = you can replace the - with = and the = with + to create an equivalent grid with only + and = in it

    so the given grid

    a+b=c

    +..-..-

    d+e=f

    =.=.=

    g-h=i

    is equivalent to

    a+b=c

    +..=..=

    d+e=f

    =.+.+

    g=h+i

    next, considering the all of the sums

    5=2+3

    6=3+3

    8=2+6=3+5

    9=3+6

    11=2+9=3+8=5+6

    8&11 must be in c or g since they are the only ones which are a result of two sums

    and 2&3 must be in a or i since they are the only ones which are a result of no sums. Due to symmetry about the diagonal gec, it doesn't matter which is which so without loss of generality say a=2 & i=3

    next 9 only exists in the sum 2+9=11 so 9 must be b or d, but if it was in d then it would be the result of two sums instead of just the one sum 3+6=9 so b=9, which means c=11, g=8

    so 3&6 must be in e or h, but h must be 5 which is a contradiction

    so there's no solution

    .,,.

  • 1 decade ago

    If we named it as

    abc

    def

    ghi

    We'll get that

    2(a+b+d) = 2(f+h+i)

    so

    a+b+d = f+h+i

    And c or g ought to be 11

    so a+b or a+d ought to be 11

    so

    11+d = f+h+i

    or 11+b = f+h+i

    so

    11 + x = f+h+i

    and we are to chose x,f,h,i from 2,3,3,5,6,8,8,9

    11+2 => no f h i

    11+3 =14 = 3+5+6 or 9+3+2

    11+5 =16 = 2+6+8

    11+6 = 17 = 9+5+3

    11+8 = 19 = 9+8+2

    11+9 = 20 = no f, h, i

    And i don’t get an answer for those values of f,h,i

    May be there is no answer.

    But its nearly 2.30am

    Tomorrow i'll try.

    Just b4 I hit the bed=>

    http://i579.photobucket.com/albums/ss239/Nino_ops/...

    Basically the guy bellow me might be is saying. But took me only a=I since I figure-out a+b+d=f+h+i

    PS: might be rubbish, (given that i was struggling to keep my eyes open)

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  • 1 decade ago

    Wow, thanks for sharing. I applied my roughly first grade level problem solving ability to the problem before I saw the heavy hitter solutions. I had arrived at the conclusion that I agreed with you that it didn't seem to be solvable, but honestly as a former educator, have an extremely high regard for text books, their high level of error checking before publishing and consequent low probability of errors.

    As I said, far better mathematic minds than mine have opined it is impossible, but would still be interested in the "final" solution, I guess the text book company's either admission of error or their solution. I am putting you as a contact and hope if you get some resolution, you'll post a new question and we can see what it was.

    I am having a little difficulty thinking this is a first grade level problem. I have tried to approach it as if there were a solution and decided as a group problem when they probably attempt brute force solutions it would probably be appropriate and fun as well as helping provide them with problem solving skills.

    As I said thanks to you for posting and thanks to all the real answerers who provided logical, systematic proofs of why it is not possible.

  • 1 decade ago

    We can write the number as

    a b a+b

    c d c+d

    a+c b-d *

    Then

    (a+b) -(c+d) = (a+c) -(b-d)

    a+b-c-d = a+ c- b+d

    2(b-c-d) = 0

    Then b = c+d and c = b - d

    (so the (1,2) entry the same as the (2,3) entry and the (2,1) entry is the same as the (3,2) entry )

    Then the configuration must be

    * 3 *

    8 * 3

    * 8 *

    Or the transpose of it.

    Note that 11 must be either on the top right corner of on the down left corner. (otherwise there should be a number that bigger than 11)

    But we run out of 8 or 3 to make 8+3 =11 or 3 + 8=11.

  • Susie
    Lv 4
    1 decade ago

    This is tough, and I'm in high school.

    I will get it eventually, hold on.

    *EDIT*

    Well assuming you can only use those numbers and only once....

    This is impossible!! I'm not exaggerating, I spent 45 minutes trying different variations and none of them have worked. If there is an answer, I'd love to know!!! :)

  • 1 decade ago

    With the given variables, this is impossible. I think there was supposed to be two "6"s instead of two "8"s. I've tried it several ways and that's the only one that keeps coming up.

  • Anonymous
    1 decade ago

    spent ages on that but it is IMPOSSIBLE which is good for you cus now you have a reason for not being able to do your childrens homework ;P

    if your any way interested i prooved its impossible

    let the boxs equal

    A,B,C

    D,E,F

    G,H,i

    respectively

    A + B = C ...............C - F = i so C = i + F

    A + B = i + F.............D + E = F

    A + B = i + D + E....... G - H = i

    A + B = G - H + D + E.....A + D = G

    A + B = A + D - H + D + E

    B + H -E = 2D..........B - E = H

    2H = 2D

    H = D ........which means H,D either equal 3,3 or 8,8

    also

    G - D = i .........comes from G - H = i

    A + D = G

    and both equations

    A + G = i + G

    A = i .............which means A,i areither 3,3 or 8,8

    A + D = G

    8 + 3 (eitherway there added) = G

    G = 11

    G - H = i

    11 - H = i

    when H or i = 3 or 8

    H has to be 8 and i has to be 3

    11 - 8 = 3

    therfore

    B - E = 8

    which means B is greater than 8, but the 11 is aready used so we cant solve it.

  • 1 decade ago

    5 + 6=11

    2 + 3= 5

    7 - 3 =4

    another trial

    5 + 6 = 11

    3 + 6 = 9

    8 - 6= 2

    Conclusion, it is impossible.

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