BDMO NATIONAL Junior 2016/05

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Math Mad Muggle
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BDMO NATIONAL Junior 2016/05

Unread post by Math Mad Muggle » Mon Jan 30, 2017 9:33 pm

i urgently need the solution.
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ahmedittihad
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Re: BDMO NATIONAL Junior 2016/05

Unread post by ahmedittihad » Mon Jan 30, 2017 9:48 pm

$P = 1000a+100b+10c+d$, $Q = 1000d+100c+10b+a$.
So, $Q-P = 999(d-a)+90(c-b)$. Now, $0<=(c-b)<=(d-a)<=9$.
Plugging in values will solve the problem now.
Frankly, my dear, I don't give a damn.

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Thamim Zahin
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Re: BDMO NATIONAL Junior 2016/05

Unread post by Thamim Zahin » Tue Jan 31, 2017 9:05 pm

Let $N = (d_3,d_2,d_1,d_0)$ mean $N = \overline{d_3d_2d_1d_0}$. For $P = (a,b,c,d)$ with $1 \leq a \leq b \leq c \leq d$ we consider the following three cases:

Case 1: $d=a$. Then $a=b=c=d$, yielding $Q=P$, which means $X = P - Q = 0$ and $Y=0$. Hence $X + Y = 0$.

Case 2: $d>a$ and $c=b$. Then

$X = Q - P = (d,b,b,a) - (a,b,b,d) = (d - a - 1,9,9,10 + a - d)$.

If $d = a+1$, then

$X + Y = 999 + 999 = 1998$,

and if $d > a+1$, then

$X + Y = (d - a - 1,9,9,10 + a - d) + (10 + a - d,9,9,d - a - 1) = (10,9,8,9)$.

In other words, $X + Y \in \{1998,10989\}$.

Case 3: $d>a$ and $c>b$. Then

$X = Q - P = (d,c,b,a) - (a,b,c,d) = (d - a, c - b - 1, b - c + 9, 10 + a - d)$,

yielding

$X + Y = (d - a, c - b - 1, b - c + 9, 10 + a - d) + (10 + a - d, b - c + 9, c - b - 1, d - a) = (10,8,9,0)$.

In other words, $X + Y = 10890$.

Conclusion:The possible values of $X+Y$ are $0, 1998, 10890,10989$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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