## BDMO national 2016, junior 10

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### BDMO national 2016, junior 10

$a, b, c, d$ are four positive integers where $a < b < c < d$ and the sum of any three of them is divisible by the fourth. Find all possible values of $(a, b, c, d)$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm

### Re: BDMO national 2016, junior 10

Note that this is equivalent to $a,b,c,d\mid a+b+c+d$ or $\text{lcm}(a,b,c,d)\mid a+b+c+d$. Letting $\text{lcm}(a,b,c,d)=X$, we want, $\dfrac{a}{X}+\dfrac{b}{X}+\dfrac{c}{X}+\dfrac{d}{X}=\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=k\in\mathbb{Z}^+$. The denominators are all integers so we have only four cases: $k=1, 2, 3, 4$ and this becomes a classic Diophantine problem...

If $k=4$ then $X/a=X/b=X/c=X/d=1$ so $a=b=c=d$ so they are not distinct, contradiction.

If $k=3$ then WLOG $X/a=X/b=1$, $X/c=X/d=2$ so $a=b$ which is also bad.

If $k=2$ then $X/a\le 2$ else $\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{4}{3} < 2$.
If $X/a=2$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{3}{2}$ where $X/b \le 2$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le 1 < \dfrac{3}{2}$. However this contradicts $X/a < X/b$ so no solutions.
Then we must have $X/a=1$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=1$ where $X/b \le 3$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{3}{4} < 1$.
If $X/b=3$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{2}{3}$ implies $2X/c\cdot X/d - 3X/c-3X/d =0$implies $(2X/c-3)(2X/d-3)=9$ which has the solutions $X/c=2, X/d=6$ which doesn't work as $X/b > X/c$ or $X/c=X/d=3$ which doesn't work as $c=d$.
Thus $X/b=2$ which gives $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{2}$ implies $(X/c-2)(X/d-2)=4$ which has solutions $(X/c, X/d)=(3, 6), (4,4)$ and the last solution doesn't work since $c=d$.
Finally $X/b \le 1$ doesn't work as $a=b$.

$k=1$ is the interesting and tedious case. WLOG $X/a\le X/b\le X/c\le X/d$, then $X/a\le 4$ else $\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{4}{5} < 1$. If $X/a=4$ then $X/b=X/c=X/d=4$ which is bad because non-distinct.
If $X/a=3$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{2}{3}$, for which $X/b\le 4$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} = \dfrac{3}{5} < \dfrac{2}{3}$.
If $X/b = 4$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d} = 5/12$ and so $5\cdot X/c\cdot X/d -12X/c-12X/d =0$ implies $(5X/c-12)(5X/d-12)=144$ and after bashing factors (eww) the only solutions are $5X/c-12=3,5X/d-12=48$ and $5X/c-12=8, 5X/d-12=18$ for which $X/c=3$ and $X/d=12$ which doesn't work as $X/a=X/c$, or $X/c=4$ and $X/d=6$ which also doesn't work as $X/b=X/c$.
If $X/b=3$ then $a=b$ which doesn't work, and if $X/b < 3$ then $X/b < X/a$ which defies the condition.
Thus $X/a=2$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{2}$ for which $X/b \le 6$, else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{3}{7} < \dfrac{1}{2}$.
If $X/b=6$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d} = 1/3$implies $X/c\cdot X/d -3X/c-3X/d=0$implies $(X/c-3)(X/d-3)=9$ which has the solutions $X/c-3=1$, $X/d-3=9$ which gives $X/c=4$, $X/d=12$ which doesn't work as $X/c < X/b$ or $X/c-3=3$, $X/d-3=3$ which doesn't work as $c=d$.
If $X/b=5$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{3}{10}$ implies $3\cdot X/c\cdot X/d - 10X/c-10X/d=0$ implies $(3X/c-10)(3X/d-10)=100$ which only has the solutions $3X/c-10 =2, 3X/d-10 = 50$ or $X/c=4$ and $X/d=20$ which doesn't contradicts $X/b < X/c$; and $3X/c-10=5, 3X/d-10=20$ which gives $X/c=5$ and $X/d=10$ which gives $b=c$ which doesn't work.
If $X/b=4$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{4}$ so $(X/c-4)(X/d-4)=16$ which has solutions $(X/c,X/d)=(5,20), (6, 12), (8,8)$ the last which gives $c=d$ which is bad.
If $X/b=3$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{6}$ so $(X/c-6)(X/d-6)=36$ which has solutions $(X/c, X/d)=(7,42), (8,24), (9, 18), (10, 15), (12, 12)$, the last of which gives $c=d$ which is bad.
$X/b \le 2$ gives contradiction with $X/b > X/a$.
$X/a=1$ is impossible as $X/b, X/c, X/d=0$ so we have exhausted all solutions.

Thus, our solutions are: $(X/a, X/b, X/c, X/d)=(1, 2, 3, 6), (2, 4, 5, 20), (2, 4, 6, 12), (2, 3, 7,42), (2, 3, 8, 24), (2, 3, 9, 18), (2, 3, 10, 15)$ for which $(2,4,6,12)$ doesn't work because it doesn't give $X$ as the LCM.

These solutions correspond to: $$\boxed{(a,b,c,d)= (6k, 3k, 2k, k),(10k, 5k, 4k, k), (21k, 14k, 6k, k), (12k, 8k, 3k, k), (9k, 6k, 2k, k), (15k, 10k, 3k, 2k)}$$ for $k\in\mathbb{Z}^+$
Frankly, my dear, I don't give a damn.

Kazi_Zareer
Posts: 86
Joined: Thu Aug 20, 2015 7:11 pm
Location: Malibagh,Dhaka-1217

### Re: BDMO national 2016, junior 10

ahmedittihad wrote:Note that this is equivalent to $a,b,c,d\mid a+b+c+d$ or $\text{lcm}(a,b,c,d)\mid a+b+c+d$. Letting $\text{lcm}(a,b,c,d)=X$, we want, $\dfrac{a}{X}+\dfrac{b}{X}+\dfrac{c}{X}+\dfrac{d}{X}=\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=k\in\mathbb{Z}^+$. The denominators are all integers so we have only four cases: $k=1, 2, 3, 4$ and this becomes a classic Diophantine problem...

If $k=4$ then $X/a=X/b=X/c=X/d=1$ so $a=b=c=d$ so they are not distinct, contradiction.

If $k=3$ then WLOG $X/a=X/b=1$, $X/c=X/d=2$ so $a=b$ which is also bad.

If $k=2$ then $X/a\le 2$ else $\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{4}{3} < 2$.
If $X/a=2$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{3}{2}$ where $X/b \le 2$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le 1 < \dfrac{3}{2}$. However this contradicts $X/a < X/b$ so no solutions.
Then we must have $X/a=1$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=1$ where $X/b \le 3$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{3}{4} < 1$.
If $X/b=3$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{2}{3}$ implies $2X/c\cdot X/d - 3X/c-3X/d =0$implies $(2X/c-3)(2X/d-3)=9$ which has the solutions $X/c=2, X/d=6$ which doesn't work as $X/b > X/c$ or $X/c=X/d=3$ which doesn't work as $c=d$.
Thus $X/b=2$ which gives $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{2}$ implies $(X/c-2)(X/d-2)=4$ which has solutions $(X/c, X/d)=(3, 6), (4,4)$ and the last solution doesn't work since $c=d$.
Finally $X/b \le 1$ doesn't work as $a=b$.

$k=1$ is the interesting and tedious case. WLOG $X/a\le X/b\le X/c\le X/d$, then $X/a\le 4$ else $\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{4}{5} < 1$. If $X/a=4$ then $X/b=X/c=X/d=4$ which is bad because non-distinct.
If $X/a=3$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{2}{3}$, for which $X/b\le 4$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} = \dfrac{3}{5} < \dfrac{2}{3}$.
If $X/b = 4$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d} = 5/12$ and so $5\cdot X/c\cdot X/d -12X/c-12X/d =0$ implies $(5X/c-12)(5X/d-12)=144$ and after bashing factors (eww) the only solutions are $5X/c-12=3,5X/d-12=48$ and $5X/c-12=8, 5X/d-12=18$ for which $X/c=3$ and $X/d=12$ which doesn't work as $X/a=X/c$, or $X/c=4$ and $X/d=6$ which also doesn't work as $X/b=X/c$.
If $X/b=3$ then $a=b$ which doesn't work, and if $X/b < 3$ then $X/b < X/a$ which defies the condition.
Thus $X/a=2$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{2}$ for which $X/b \le 6$, else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{3}{7} < \dfrac{1}{2}$.
If $X/b=6$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d} = 1/3$implies $X/c\cdot X/d -3X/c-3X/d=0$implies $(X/c-3)(X/d-3)=9$ which has the solutions $X/c-3=1$, $X/d-3=9$ which gives $X/c=4$, $X/d=12$ which doesn't work as $X/c < X/b$ or $X/c-3=3$, $X/d-3=3$ which doesn't work as $c=d$.
If $X/b=5$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{3}{10}$ implies $3\cdot X/c\cdot X/d - 10X/c-10X/d=0$ implies $(3X/c-10)(3X/d-10)=100$ which only has the solutions $3X/c-10 =2, 3X/d-10 = 50$ or $X/c=4$ and $X/d=20$ which doesn't contradicts $X/b < X/c$; and $3X/c-10=5, 3X/d-10=20$ which gives $X/c=5$ and $X/d=10$ which gives $b=c$ which doesn't work.
If $X/b=4$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{4}$ so $(X/c-4)(X/d-4)=16$ which has solutions $(X/c,X/d)=(5,20), (6, 12), (8,8)$ the last which gives $c=d$ which is bad.
If $X/b=3$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{6}$ so $(X/c-6)(X/d-6)=36$ which has solutions $(X/c, X/d)=(7,42), (8,24), (9, 18), (10, 15), (12, 12)$, the last of which gives $c=d$ which is bad.
$X/b \le 2$ gives contradiction with $X/b > X/a$.
$X/a=1$ is impossible as $X/b, X/c, X/d=0$ so we have exhausted all solutions.

Thus, our solutions are: $(X/a, X/b, X/c, X/d)=(1, 2, 3, 6), (2, 4, 5, 20), (2, 4, 6, 12), (2, 3, 7,42), (2, 3, 8, 24), (2, 3, 9, 18), (2, 3, 10, 15)$ for which $(2,4,6,12)$ doesn't work because it doesn't give $X$ as the LCM.

These solutions correspond to: $$\boxed{(a,b,c,d)= (6k, 3k, 2k, k),(10k, 5k, 4k, k), (21k, 14k, 6k, k), (12k, 8k, 3k, k), (9k, 6k, 2k, k), (15k, 10k, 3k, 2k)}$$ for $k\in\mathbb{Z}^+$
Frankly, my dear, did you type this whole solution?! Great patience indeed, man!
We cannot solve our problems with the same thinking we used when we create them.

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### Re: BDMO national 2016, junior 10

ahmedittihad wrote:Note that this is equivalent to $a,b,c,d\mid a+b+c+d$ or $\text{lcm}(a,b,c,d)\mid a+b+c+d$. Letting $\text{lcm}(a,b,c,d)=X$, we want, $\dfrac{a}{X}+\dfrac{b}{X}+\dfrac{c}{X}+\dfrac{d}{X}=\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=k\in\mathbb{Z}^+$. The denominators are all integers so we have only four cases: $k=1, 2, 3, 4$ and this becomes a classic Diophantine problem...

If $k=4$ then $X/a=X/b=X/c=X/d=1$ so $a=b=c=d$ so they are not distinct, contradiction.

If $k=3$ then WLOG $X/a=X/b=1$, $X/c=X/d=2$ so $a=b$ which is also bad.

If $k=2$ then $X/a\le 2$ else $\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{4}{3} < 2$.
If $X/a=2$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{3}{2}$ where $X/b \le 2$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le 1 < \dfrac{3}{2}$. However this contradicts $X/a < X/b$ so no solutions.
Then we must have $X/a=1$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=1$ where $X/b \le 3$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{3}{4} < 1$.
If $X/b=3$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{2}{3}$ implies $2X/c\cdot X/d - 3X/c-3X/d =0$implies $(2X/c-3)(2X/d-3)=9$ which has the solutions $X/c=2, X/d=6$ which doesn't work as $X/b > X/c$ or $X/c=X/d=3$ which doesn't work as $c=d$.
Thus $X/b=2$ which gives $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{2}$ implies $(X/c-2)(X/d-2)=4$ which has solutions $(X/c, X/d)=(3, 6), (4,4)$ and the last solution doesn't work since $c=d$.
Finally $X/b \le 1$ doesn't work as $a=b$.

$k=1$ is the interesting and tedious case. WLOG $X/a\le X/b\le X/c\le X/d$, then $X/a\le 4$ else $\dfrac{1}{X/a}+\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{4}{5} < 1$. If $X/a=4$ then $X/b=X/c=X/d=4$ which is bad because non-distinct.
If $X/a=3$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{2}{3}$, for which $X/b\le 4$ else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} = \dfrac{3}{5} < \dfrac{2}{3}$.
If $X/b = 4$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d} = 5/12$ and so $5\cdot X/c\cdot X/d -12X/c-12X/d =0$ implies $(5X/c-12)(5X/d-12)=144$ and after bashing factors (eww) the only solutions are $5X/c-12=3,5X/d-12=48$ and $5X/c-12=8, 5X/d-12=18$ for which $X/c=3$ and $X/d=12$ which doesn't work as $X/a=X/c$, or $X/c=4$ and $X/d=6$ which also doesn't work as $X/b=X/c$.
If $X/b=3$ then $a=b$ which doesn't work, and if $X/b < 3$ then $X/b < X/a$ which defies the condition.
Thus $X/a=2$ then $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{2}$ for which $X/b \le 6$, else $\dfrac{1}{X/b}+\dfrac{1}{X/c}+\dfrac{1}{X/d} \le \dfrac{3}{7} < \dfrac{1}{2}$.
If $X/b=6$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d} = 1/3$implies $X/c\cdot X/d -3X/c-3X/d=0$implies $(X/c-3)(X/d-3)=9$ which has the solutions $X/c-3=1$, $X/d-3=9$ which gives $X/c=4$, $X/d=12$ which doesn't work as $X/c < X/b$ or $X/c-3=3$, $X/d-3=3$ which doesn't work as $c=d$.
If $X/b=5$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{3}{10}$ implies $3\cdot X/c\cdot X/d - 10X/c-10X/d=0$ implies $(3X/c-10)(3X/d-10)=100$ which only has the solutions $3X/c-10 =2, 3X/d-10 = 50$ or $X/c=4$ and $X/d=20$ which doesn't contradicts $X/b < X/c$; and $3X/c-10=5, 3X/d-10=20$ which gives $X/c=5$ and $X/d=10$ which gives $b=c$ which doesn't work.
If $X/b=4$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{4}$ so $(X/c-4)(X/d-4)=16$ which has solutions $(X/c,X/d)=(5,20), (6, 12), (8,8)$ the last which gives $c=d$ which is bad.
If $X/b=3$ then $\dfrac{1}{X/c}+\dfrac{1}{X/d}=\dfrac{1}{6}$ so $(X/c-6)(X/d-6)=36$ which has solutions $(X/c, X/d)=(7,42), (8,24), (9, 18), (10, 15), (12, 12)$, the last of which gives $c=d$ which is bad.
$X/b \le 2$ gives contradiction with $X/b > X/a$.
$X/a=1$ is impossible as $X/b, X/c, X/d=0$ so we have exhausted all solutions.

Thus, our solutions are: $(X/a, X/b, X/c, X/d)=(1, 2, 3, 6), (2, 4, 5, 20), (2, 4, 6, 12), (2, 3, 7,42), (2, 3, 8, 24), (2, 3, 9, 18), (2, 3, 10, 15)$ for which $(2,4,6,12)$ doesn't work because it doesn't give $X$ as the LCM.

These solutions correspond to: $$\boxed{(a,b,c,d)= (6k, 3k, 2k, k),(10k, 5k, 4k, k), (21k, 14k, 6k, k), (12k, 8k, 3k, k), (9k, 6k, 2k, k), (15k, 10k, 3k, 2k)}$$ for $k\in\mathbb{Z}^+$
Thanks that you took your time and wrote down the whole solution. A question, how long did it take?
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm

### Re: BDMO national 2016, junior 10

I knew this problem was pretty much bland and disgusting. So, I just copied the solution from AOPS. Just copying took me 15 minutes. Allah knows how much time it took for the person to write.
Frankly, my dear, I don't give a damn.

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm

### Re: BDMO national 2016, junior 10

ahmedittihad wrote:I knew this problem was pretty much bland and disgusting. So, I just copied the solution from AOPS. Just copying took me 15 minutes. Allah knows how much time it took for the person to write.
Cheeky
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

Kazi_Zareer
Posts: 86
Joined: Thu Aug 20, 2015 7:11 pm
Location: Malibagh,Dhaka-1217

### Re: BDMO national 2016, junior 10

ahmedittihad wrote:I knew this problem was pretty much bland and disgusting. So, I just copied the solution from AOPS. Just copying took me 15 minutes. Allah knows how much time it took for the person to write.
Haha! It may be a well known NT problem. So sad that in bdmo national 2016 Junior Category's some problems(especially, The art gallery problem) were given from various well known resources.
We cannot solve our problems with the same thinking we used when we create them.

prottoydas
Posts: 8
Joined: Thu Feb 01, 2018 11:56 am