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Thread: Brain Teaser Help Please~!

  1. #1
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    Brain Teaser Help Please~!

    Hi guys, I have been solving some problems recently, but there are few problems on which i am stuck:

    1) If (a/b)-a=6 what is the value of sqrt of ((a+ab-2b)/b)?

    2) For positive interger p,t,x, and y, if p^x=t^y and x-y=3, which of the follwoing cannot equal t?

    a. 1 b.2 c.4 d. 9 e. 25

    3)If a,b,c are distinct positive intergers and 10% of abc is 5, then a+b could equal...

    a.1 b.3 c.5 d.6 e.25

    I fully appreciate your help!
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  2. #2
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    for 3). if 10% of abc is 5 then it means that abc = 50.

    Hence, a,b and c must all be divisible by 50 (assuming whole numbers only)

    a+b cant be 1 because we are talking only positive integers (see question).

    if a+b equals 3, the either a=1 and b=2 or a=2 and b=1. If this was the case then ab would equal 2. Hence c could be 25. This is a possibility

    if a+b=5, then combinations are (1,4), (2,3), (3,2), (4,1). If ab was 4, then c cant be a whole number so we can discard. if ab was 6, then c cant be a whole number so we can discard.

    if a+b=6 the combinations are (1,5), (2,4), (3,3), (4,2), (5,1). if ab=5, then c could be 10. Hence a possibility. if ab=8, the not divisible by 50. if ab=9 then not divisible.

    if a+b=25, then ab could numerous combinations. Will let you work this out.

    Hence 3 and 6 are possibilities
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  3. #3
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    Hello, jjkins!

    3) If $\displaystyle a,b,c$ are distinct positive intergers,
    and 10% of $\displaystyle abc$ is 5, then $\displaystyle a+b$ could equal...

    . . $\displaystyle a)\;1\qquad b)\;3\qquad c)\;5\qquad d)\;6\qquad e)\;25$

    If $\displaystyle 10\% \times abc = 5$, then: .$\displaystyle abc \:=\:50.$

    Since $\displaystyle a,b,c$ are distinct positive integers, $\displaystyle a,b,c \:=\:\{1,2,25\}\:\text{ or }\:\{1,5,10\}$

    The possible sums are: .$\displaystyle \begin{array}{ccc}1 + 2 &=&{\color{blue}3} \\ 1 + 25 &=&26 \\ 2 + 25 &=& 27\end{array}\qquad\begin{array}{ccc}1+5 &=&{\color{blue}6} \\ 1+10 &=&11 \\ 5+10 &=&15 \end{array}$


    There are two possible answers: .3 and 6.

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  4. #4
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    Quote Originally Posted by jjkins View Post
    ...there are few problems on which i am stuck:

    1) If (a/b)-a=6 [I]
    what is the value of sqrt((a+ab-2b)/b) [II]?

    ...
    [I] : $\displaystyle \frac ab-a=6~\implies~ \frac ab = a+6$

    [II] : $\displaystyle \sqrt{\frac{a+ab-2b}{b}} = \sqrt{\frac ab +a - 2}$

    Plug in the term for $\displaystyle \frac ab$ into the term [II] and simplify:

    $\displaystyle \sqrt{a+6 +a - 2} = \boxed{\sqrt{2a+4}}$
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  5. #5
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by jjkins View Post
    2) For positive interger p,t,x, and y, if p^x=t^y and x-y=3, which of the follwoing cannot equal t?

    a. 1 b.2 c.4 d. 9 e. 25
    If $\displaystyle t=2$ then $\displaystyle p^x=2^y$; since 2 is prime and all the variables are positive integers, we must have $\displaystyle p=2^n$ for some positive integer n. $\displaystyle \therefore\ xn=y$ $\displaystyle \Rightarrow$ $\displaystyle 3n+3ny=y$ $\displaystyle \Rightarrow$ $\displaystyle 3n=(1-3n)y$ $\displaystyle \Rightarrow$ $\displaystyle y=\frac{3n}{1-3n}$ which is not positive or an integer for $\displaystyle n\ge1$.

    Hence it is not possible for t to be 2.

    On the other handm setting $\displaystyle x=6,y=3$ yields $\displaystyle p^{x}=t^y$ $\displaystyle \Rightarrow$ $\displaystyle p^6=t^3$ $\displaystyle \Rightarrow$ $\displaystyle p^2=t$ so it is possible for t to be a perfect square (i.e. 1, 4, 9 and 25 are all possible answers).
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