BDMO NATIONAL Junior 2016/05

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BDMO NATIONAL Junior 2016/05
i urgently need the solution.
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 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO NATIONAL Junior 2016/05
$P = 1000a+100b+10c+d$, $Q = 1000d+100c+10b+a$.
So, $QP = 999(da)+90(cb)$. Now, $0<=(cb)<=(da)<=9$.
Plugging in values will solve the problem now.
So, $QP = 999(da)+90(cb)$. Now, $0<=(cb)<=(da)<=9$.
Plugging in values will solve the problem now.
Frankly, my dear, I don't give a damn.
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: BDMO NATIONAL Junior 2016/05
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 $0, 1998, 10890,10989$.
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.