Mixing
Geometric and arithmetic mean
Clash Royale CLAN TAG#URR8PPP
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The positive integers $a<b$ are such that
$fraca+b2$
and $sqrtab$ are positive
integers consisting of the same two digits in reverse order. What is the minimum
value of a?
The first thing that comes into mind is trial and error as there are just 100 ways, but I need a fast way to do it.
I tried letting $10x + y$ and $10y + x$ be the 2 numbers
from the 2 equations,
$fraca+b2 = 10x + y$
$sqrtab = 10y +x$
$100y^2 +10xy +x^2 = 20ax +2ay-a^2$
A quadratic equation in $a$
$a^2 -2ay-20ax+100y^2+10xy+x^2$
but getting the discriminant and finding integer solutions is really tedious.
optimization
asked Jul 21 at 6:31
SuperMage1
680210
add a comment |Â
up vote
0
down vote
favorite
The positive integers $a<b$ are such that
$fraca+b2$
and $sqrtab$ are positive
integers consisting of the same two digits in reverse order. What is the minimum
value of a?
The first thing that comes into mind is trial and error as there are just 100 ways, but I need a fast way to do it.
I tried letting $10x + y$ and $10y + x$ be the 2 numbers
from the 2 equations,
$fraca+b2 = 10x + y$
$sqrtab = 10y +x$
$100y^2 +10xy +x^2 = 20ax +2ay-a^2$
A quadratic equation in $a$
$a^2 -2ay-20ax+100y^2+10xy+x^2$
but getting the discriminant and finding integer solutions is really tedious.
optimization
asked Jul 21 at 6:31
SuperMage1
680210
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The positive integers $a<b$ are such that
$fraca+b2$
and $sqrtab$ are positive
integers consisting of the same two digits in reverse order. What is the minimum
value of a?
The first thing that comes into mind is trial and error as there are just 100 ways, but I need a fast way to do it.
I tried letting $10x + y$ and $10y + x$ be the 2 numbers
from the 2 equations,
$fraca+b2 = 10x + y$
$sqrtab = 10y +x$
$100y^2 +10xy +x^2 = 20ax +2ay-a^2$
A quadratic equation in $a$
$a^2 -2ay-20ax+100y^2+10xy+x^2$
but getting the discriminant and finding integer solutions is really tedious.
optimization
asked Jul 21 at 6:31
SuperMage1
680210
The positive integers $a<b$ are such that
$fraca+b2$
and $sqrtab$ are positive
integers consisting of the same two digits in reverse order. What is the minimum
value of a?
The first thing that comes into mind is trial and error as there are just 100 ways, but I need a fast way to do it.
I tried letting $10x + y$ and $10y + x$ be the 2 numbers
from the 2 equations,
$fraca+b2 = 10x + y$
$sqrtab = 10y +x$
$100y^2 +10xy +x^2 = 20ax +2ay-a^2$
A quadratic equation in $a$
$a^2 -2ay-20ax+100y^2+10xy+x^2$
but getting the discriminant and finding integer solutions is really tedious.
optimization
asked Jul 21 at 6:31
SuperMage1
680210
asked Jul 21 at 6:31
SuperMage1
680210
asked Jul 21 at 6:31
SuperMage1
680210
asked Jul 21 at 6:31
SuperMage1
680210
680210
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
$$fraca+b2=10x+y implies a^2+2ab+b^2=2^2(10x+y)^2$$
$$sqrtab=10y+x implies 4ab=4(10y+x)^2$$
Subtracting the equations give us
$$(a-b)^2=4(9x-9y)(11x+11y)=4(9)(11)(x-y)(x+y)=4(9)(11)(x^2-y^2)$$
Hence there exists $nin mathbbZ^+$, such that $(x-y)(x+y)=11n^2$.
We know that $x le 9$, $ 11n^2=x^2-y^2le 81 implies n^2 le frac8111<9$, hence $n^2 in 1,4$.
If $n^2=4$, then we have $$(x+y)(x-y)=44$$
Let's consider the factorization of $44$.
beginalign
44&=1 times 44 \
&=2 times 22 \
&= 4 times 11
endalign
Since $x+y le 18$, then we must have $x+y = 11$, and $x-y=4$. However, this would lead to $2x=15$.
Hence, we must have $n^2=1$.
$$x+y=11, x-y=1$$
That is $x=6, y=5$.
$$fraca+b2=65, sqrtab=56$$
$$(a-b)^2=4(9)(11)(x^2-y^2)=(2cdot 3 cdot 11)^2$$
$$a-b=-66$$
$$a+b=130$$
Hence $a=frac642=32.$
answered Jul 21 at 7:54
Siong Thye Goh
77.6k134795
1
You forgot that $a<b$, but the fix is immediate.
– Yves Daoust
Jul 21 at 8:27
thank you, I indeed forgot which is bigger.
– Siong Thye Goh
Jul 21 at 12:56
add a comment |Â
up vote
1
down vote
$$a+b=20d+2u,\ab=(10u+d)^2,$$
implies that
$$(b-a)^2=(20d+2u)^2-4(10u+d)^2=6^2,11,(d+u)(d-u)$$ must be a perfect square. Then $d+u$ must be $11$, while $d-u$ must be a odd perfect square (i.e. one of $1,9$, but $9$ doesn't work).
The only match is $65$, giving $a+b=130,a-b=66$ then $a=32,b=98$.
answered Jul 21 at 8:15
Yves Daoust
111k665204
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$$fraca+b2=10x+y implies a^2+2ab+b^2=2^2(10x+y)^2$$
$$sqrtab=10y+x implies 4ab=4(10y+x)^2$$
Subtracting the equations give us
$$(a-b)^2=4(9x-9y)(11x+11y)=4(9)(11)(x-y)(x+y)=4(9)(11)(x^2-y^2)$$
Hence there exists $nin mathbbZ^+$, such that $(x-y)(x+y)=11n^2$.
We know that $x le 9$, $ 11n^2=x^2-y^2le 81 implies n^2 le frac8111<9$, hence $n^2 in 1,4$.
If $n^2=4$, then we have $$(x+y)(x-y)=44$$
Let's consider the factorization of $44$.
beginalign
44&=1 times 44 \
&=2 times 22 \
&= 4 times 11
endalign
Since $x+y le 18$, then we must have $x+y = 11$, and $x-y=4$. However, this would lead to $2x=15$.
Hence, we must have $n^2=1$.
$$x+y=11, x-y=1$$
That is $x=6, y=5$.
$$fraca+b2=65, sqrtab=56$$
$$(a-b)^2=4(9)(11)(x^2-y^2)=(2cdot 3 cdot 11)^2$$
$$a-b=-66$$
$$a+b=130$$
Hence $a=frac642=32.$
answered Jul 21 at 7:54
Siong Thye Goh
77.6k134795
1
You forgot that $a<b$, but the fix is immediate.
– Yves Daoust
Jul 21 at 8:27
thank you, I indeed forgot which is bigger.
– Siong Thye Goh
Jul 21 at 12:56
add a comment |Â
up vote
3
down vote
$$fraca+b2=10x+y implies a^2+2ab+b^2=2^2(10x+y)^2$$
$$sqrtab=10y+x implies 4ab=4(10y+x)^2$$
Subtracting the equations give us
$$(a-b)^2=4(9x-9y)(11x+11y)=4(9)(11)(x-y)(x+y)=4(9)(11)(x^2-y^2)$$
Hence there exists $nin mathbbZ^+$, such that $(x-y)(x+y)=11n^2$.
We know that $x le 9$, $ 11n^2=x^2-y^2le 81 implies n^2 le frac8111<9$, hence $n^2 in 1,4$.
If $n^2=4$, then we have $$(x+y)(x-y)=44$$
Let's consider the factorization of $44$.
beginalign
44&=1 times 44 \
&=2 times 22 \
&= 4 times 11
endalign
Since $x+y le 18$, then we must have $x+y = 11$, and $x-y=4$. However, this would lead to $2x=15$.
Hence, we must have $n^2=1$.
$$x+y=11, x-y=1$$
That is $x=6, y=5$.
$$fraca+b2=65, sqrtab=56$$
$$(a-b)^2=4(9)(11)(x^2-y^2)=(2cdot 3 cdot 11)^2$$
$$a-b=-66$$
$$a+b=130$$
Hence $a=frac642=32.$
answered Jul 21 at 7:54
Siong Thye Goh
77.6k134795
1
You forgot that $a<b$, but the fix is immediate.
– Yves Daoust
Jul 21 at 8:27
thank you, I indeed forgot which is bigger.
– Siong Thye Goh
Jul 21 at 12:56
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$fraca+b2=10x+y implies a^2+2ab+b^2=2^2(10x+y)^2$$
$$sqrtab=10y+x implies 4ab=4(10y+x)^2$$
Subtracting the equations give us
$$(a-b)^2=4(9x-9y)(11x+11y)=4(9)(11)(x-y)(x+y)=4(9)(11)(x^2-y^2)$$
Hence there exists $nin mathbbZ^+$, such that $(x-y)(x+y)=11n^2$.
We know that $x le 9$, $ 11n^2=x^2-y^2le 81 implies n^2 le frac8111<9$, hence $n^2 in 1,4$.
If $n^2=4$, then we have $$(x+y)(x-y)=44$$
Let's consider the factorization of $44$.
beginalign
44&=1 times 44 \
&=2 times 22 \
&= 4 times 11
endalign
Since $x+y le 18$, then we must have $x+y = 11$, and $x-y=4$. However, this would lead to $2x=15$.
Hence, we must have $n^2=1$.
$$x+y=11, x-y=1$$
That is $x=6, y=5$.
$$fraca+b2=65, sqrtab=56$$
$$(a-b)^2=4(9)(11)(x^2-y^2)=(2cdot 3 cdot 11)^2$$
$$a-b=-66$$
$$a+b=130$$
Hence $a=frac642=32.$
answered Jul 21 at 7:54
Siong Thye Goh
77.6k134795
$$fraca+b2=10x+y implies a^2+2ab+b^2=2^2(10x+y)^2$$
$$sqrtab=10y+x implies 4ab=4(10y+x)^2$$
Subtracting the equations give us
$$(a-b)^2=4(9x-9y)(11x+11y)=4(9)(11)(x-y)(x+y)=4(9)(11)(x^2-y^2)$$
Hence there exists $nin mathbbZ^+$, such that $(x-y)(x+y)=11n^2$.
We know that $x le 9$, $ 11n^2=x^2-y^2le 81 implies n^2 le frac8111<9$, hence $n^2 in 1,4$.
If $n^2=4$, then we have $$(x+y)(x-y)=44$$
Let's consider the factorization of $44$.
beginalign
44&=1 times 44 \
&=2 times 22 \
&= 4 times 11
endalign
Since $x+y le 18$, then we must have $x+y = 11$, and $x-y=4$. However, this would lead to $2x=15$.
Hence, we must have $n^2=1$.
$$x+y=11, x-y=1$$
That is $x=6, y=5$.
$$fraca+b2=65, sqrtab=56$$
$$(a-b)^2=4(9)(11)(x^2-y^2)=(2cdot 3 cdot 11)^2$$
$$a-b=-66$$
$$a+b=130$$
Hence $a=frac642=32.$
answered Jul 21 at 7:54
Siong Thye Goh
77.6k134795
edited Jul 21 at 12:56
answered Jul 21 at 7:54
Siong Thye Goh
77.6k134795
answered Jul 21 at 7:54
Siong Thye Goh
77.6k134795
answered Jul 21 at 7:54
Siong Thye Goh
77.6k134795
77.6k134795
1
You forgot that $a<b$, but the fix is immediate.
– Yves Daoust
Jul 21 at 8:27
thank you, I indeed forgot which is bigger.
– Siong Thye Goh
Jul 21 at 12:56
add a comment |Â
1
You forgot that $a<b$, but the fix is immediate.
– Yves Daoust
Jul 21 at 8:27
thank you, I indeed forgot which is bigger.
– Siong Thye Goh
Jul 21 at 12:56
1
1
You forgot that $a<b$, but the fix is immediate.
– Yves Daoust
Jul 21 at 8:27
You forgot that $a<b$, but the fix is immediate.
– Yves Daoust
Jul 21 at 8:27
thank you, I indeed forgot which is bigger.
– Siong Thye Goh
Jul 21 at 12:56
thank you, I indeed forgot which is bigger.
– Siong Thye Goh
Jul 21 at 12:56
add a comment |Â
up vote
1
down vote
$$a+b=20d+2u,\ab=(10u+d)^2,$$
implies that
$$(b-a)^2=(20d+2u)^2-4(10u+d)^2=6^2,11,(d+u)(d-u)$$ must be a perfect square. Then $d+u$ must be $11$, while $d-u$ must be a odd perfect square (i.e. one of $1,9$, but $9$ doesn't work).
The only match is $65$, giving $a+b=130,a-b=66$ then $a=32,b=98$.
answered Jul 21 at 8:15
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
$$a+b=20d+2u,\ab=(10u+d)^2,$$
implies that
$$(b-a)^2=(20d+2u)^2-4(10u+d)^2=6^2,11,(d+u)(d-u)$$ must be a perfect square. Then $d+u$ must be $11$, while $d-u$ must be a odd perfect square (i.e. one of $1,9$, but $9$ doesn't work).
The only match is $65$, giving $a+b=130,a-b=66$ then $a=32,b=98$.
answered Jul 21 at 8:15
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$a+b=20d+2u,\ab=(10u+d)^2,$$
implies that
$$(b-a)^2=(20d+2u)^2-4(10u+d)^2=6^2,11,(d+u)(d-u)$$ must be a perfect square. Then $d+u$ must be $11$, while $d-u$ must be a odd perfect square (i.e. one of $1,9$, but $9$ doesn't work).
The only match is $65$, giving $a+b=130,a-b=66$ then $a=32,b=98$.
answered Jul 21 at 8:15
Yves Daoust
111k665204
$$a+b=20d+2u,\ab=(10u+d)^2,$$
implies that
$$(b-a)^2=(20d+2u)^2-4(10u+d)^2=6^2,11,(d+u)(d-u)$$ must be a perfect square. Then $d+u$ must be $11$, while $d-u$ must be a odd perfect square (i.e. one of $1,9$, but $9$ doesn't work).
The only match is $65$, giving $a+b=130,a-b=66$ then $a=32,b=98$.
answered Jul 21 at 8:15
Yves Daoust
111k665204
edited Jul 21 at 8:41
answered Jul 21 at 8:15
Yves Daoust
111k665204
answered Jul 21 at 8:15
Yves Daoust
111k665204
answered Jul 21 at 8:15
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
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