remainder and min max

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buoyant Master | Next Rank: 500 Posts Default Avatar
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remainder and min max

Post Mon Jan 19, 2015 12:30 pm
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    x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
    A. 12
    B. 20
    C. 24
    D. 29
    E. 33

    OA:B

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    Marty Murray Legendary Member
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    Post Mon Jan 19, 2015 1:19 pm
    buoyant wrote:
    x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
    A. 12
    B. 20
    C. 24
    D. 29
    E. 33

    OA:B
    Well, x has to be a least 3 for a remainder of 3 when x is divided by y.

    So going to say x = 3.

    y has to be greater than 3 for x/y to generate a remainder of 3, and y has to be at least 8 for y/z to generate a remainder of 8.

    So going to say y = 8.

    Z has to be greater than 8 for y/z to generate a remainder of 8. Lowest possible integer greater than 8 is 9.

    So going with z = 9.

    x + y + z = 3 + 8 + 9 = 20

    Choose B.

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    Post Mon Jan 19, 2015 1:24 pm
    buoyant wrote:
    x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
    A. 12
    B. 20
    C. 24
    D. 29
    E. 33
    When y is divided by z, the remainder is 8.
    In other words, y is 8 more than a multiple of z:
    y = az + 8, where a is a nonnegative integer.

    The smallest possible value of y occurs when a=0:
    y = 0*z + 8 = 8.

    If y=8, then the statement above becomes:
    When y=8 is divided by z, the remainder is 8.

    For 8/z to have a remainder of 8, z must be GREATER than 8.
    To illustrate:
    If z=1, then 8/z = 8/1 = 8 R0.
    If z=2, then 8/z = 8/2 = 4 R0.
    If z=3, then 8/z = 8/3 = 2 R2.
    If z=4, then 8/z = 8/4 = 2 R0.
    If z=5, then 8/z = 8/5 = 1 R3.
    If z=6, then 8/z = 8/6 = 1 R2.
    If z=7, then 8/z = 8/7 = 1 R1.
    If z=8, then 8/z = 8/8 = 1 R0.
    If z=9, then y/z = 8/9 = 0 R8.
    Thus, the smallest possible value of z = 9.

    When x is divided by y, the remainder is 3.
    In other words, x is 3 more than a multiple of y:
    x = by + 3, where b is a nonnegative integer.

    The smallest possible value of x occurs when b=0
    x = 0*y + 3 = 3.

    Thus:
    Smallest possible value of x+y+z = 3+8+9 = 20.

    The correct answer is B.

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    Post Mon Jan 19, 2015 1:44 pm
    buoyant wrote:
    x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
    A. 12
    B. 20
    C. 24
    D. 29
    When it comes to remainders, we have a nice rule that says:

    If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.

    For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

    -------------------------

    When y is divided by z the remainder is 8
    According to the above rule, the possible values of y are: 8, 8+z, 8+2z, 8+3z...
    We want to MINIMIZE the value of y, so y = 8

    When x is divided by y the remainder is 3
    Now that we know that y = 8, we can write: When x is divided by 8 the remainder is 3
    According to the above rule, the possible values of x are: 3, 3 + 8, 3 + 2(8)...
    We want to MINIMIZE the value of x, so x = 3

    What about z????
    We already know that, when y is divided by z, the remainder is 8
    There's a rule that says, when positive integer N is divided by positive integer D, the remainder R is such that 0 < R < D
    So, for the given information, we can conclude that 0 < 8 < z
    We want to MINIMIZE the value of z, so z = 9

    So, the smallest possible value of x+y+z =3 + 8 + 9
    = 20
    = B

    Cheers,
    Brent

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    Post Sun Jan 25, 2015 4:28 pm
    Marty Murray wrote:
    Well, x has to be a least 3 for a remainder of 3 when x is divided by y.

    So going to say x = 3.
    Well ... this is mathematical Catch-22, where anyone who could follow it would already know what to do. For the perplexed, what Marty is doing here is exploiting the fact that if we're dividing integers in the remainder system, we have

    x/y = 0, with remainder x

    if y > x > 0. For instance, 3/5 = 0, with remainder 3; 4/7 = 0, with remainder 4, etc. So x is at least 3 greater than a multiple of y, and since we want the minimum value of x, we can make x = 3 and "a multiple of y" = 0, since 0 is a multiple of any integer.

    This is worth delving into a bit, as most students will struggle with it.

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