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Finding numbers that double when you switch the first and last digit
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So, I was watching this video by MindYourDecisions about finding the smallest number that doubles when you move the last digit to become the first digit. Actually, I just saw the title of the video, and I found it interesting, so I began to work on a solution. But, in fact, I had misinterpreted the question posed by the video.
I was looking for a number that doubles when you switch the first and last digits of a number. I started off my search by writing equations like:
$$ 2(10a + b) = 10b + a $$
This simplifies to:
$$ 19a = 8b$$
Since $a$ and $b$ should be integers between 0 and 9 inclusive, I figured out that there are no solutions amongst two-digit numbers.
Next, I applied this technique to three and four digit numbers, getting this equation for three digit numbers (using the convention that $a$ is the first digit, $b$ is the second, and so on):
$$ 199a + 10b = 98c $$ And for four-digit numbers,
$$ 1999a + 100b + 10c = 998d $$
Next, I made tables varying the value of $a$ and the last digit between 0 and 9, looking for numbers that were close enough so that the smaller terms in the middle could make up the difference.
I couldn't find any solutions; however, I was able to find near misses such as 37, 397, 3997, or generally, a 3 followed by a string of 9s followed by a 7, which all differ by just one.
I am wondering if there actually are any solutions to this problem, and if not, how you would go about proving that.
Thank you!
number-theory linear-algebra
asked Jul 23 at 4:30
guy600
253
migrated from mathoverflow.net Jul 23 at 5:39
This question came from our site for professional mathematicians.
add a comment |Â
up vote
1
down vote
favorite
So, I was watching this video by MindYourDecisions about finding the smallest number that doubles when you move the last digit to become the first digit. Actually, I just saw the title of the video, and I found it interesting, so I began to work on a solution. But, in fact, I had misinterpreted the question posed by the video.
I was looking for a number that doubles when you switch the first and last digits of a number. I started off my search by writing equations like:
$$ 2(10a + b) = 10b + a $$
This simplifies to:
$$ 19a = 8b$$
Since $a$ and $b$ should be integers between 0 and 9 inclusive, I figured out that there are no solutions amongst two-digit numbers.
Next, I applied this technique to three and four digit numbers, getting this equation for three digit numbers (using the convention that $a$ is the first digit, $b$ is the second, and so on):
$$ 199a + 10b = 98c $$ And for four-digit numbers,
$$ 1999a + 100b + 10c = 998d $$
Next, I made tables varying the value of $a$ and the last digit between 0 and 9, looking for numbers that were close enough so that the smaller terms in the middle could make up the difference.
I couldn't find any solutions; however, I was able to find near misses such as 37, 397, 3997, or generally, a 3 followed by a string of 9s followed by a 7, which all differ by just one.
I am wondering if there actually are any solutions to this problem, and if not, how you would go about proving that.
Thank you!
number-theory linear-algebra
asked Jul 23 at 4:30
guy600
253
migrated from mathoverflow.net Jul 23 at 5:39
This question came from our site for professional mathematicians.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So, I was watching this video by MindYourDecisions about finding the smallest number that doubles when you move the last digit to become the first digit. Actually, I just saw the title of the video, and I found it interesting, so I began to work on a solution. But, in fact, I had misinterpreted the question posed by the video.
I was looking for a number that doubles when you switch the first and last digits of a number. I started off my search by writing equations like:
$$ 2(10a + b) = 10b + a $$
This simplifies to:
$$ 19a = 8b$$
Since $a$ and $b$ should be integers between 0 and 9 inclusive, I figured out that there are no solutions amongst two-digit numbers.
Next, I applied this technique to three and four digit numbers, getting this equation for three digit numbers (using the convention that $a$ is the first digit, $b$ is the second, and so on):
$$ 199a + 10b = 98c $$ And for four-digit numbers,
$$ 1999a + 100b + 10c = 998d $$
Next, I made tables varying the value of $a$ and the last digit between 0 and 9, looking for numbers that were close enough so that the smaller terms in the middle could make up the difference.
I couldn't find any solutions; however, I was able to find near misses such as 37, 397, 3997, or generally, a 3 followed by a string of 9s followed by a 7, which all differ by just one.
I am wondering if there actually are any solutions to this problem, and if not, how you would go about proving that.
Thank you!
number-theory linear-algebra
asked Jul 23 at 4:30
guy600
253
So, I was watching this video by MindYourDecisions about finding the smallest number that doubles when you move the last digit to become the first digit. Actually, I just saw the title of the video, and I found it interesting, so I began to work on a solution. But, in fact, I had misinterpreted the question posed by the video.
I was looking for a number that doubles when you switch the first and last digits of a number. I started off my search by writing equations like:
$$ 2(10a + b) = 10b + a $$
This simplifies to:
$$ 19a = 8b$$
Since $a$ and $b$ should be integers between 0 and 9 inclusive, I figured out that there are no solutions amongst two-digit numbers.
Next, I applied this technique to three and four digit numbers, getting this equation for three digit numbers (using the convention that $a$ is the first digit, $b$ is the second, and so on):
$$ 199a + 10b = 98c $$ And for four-digit numbers,
$$ 1999a + 100b + 10c = 998d $$
Next, I made tables varying the value of $a$ and the last digit between 0 and 9, looking for numbers that were close enough so that the smaller terms in the middle could make up the difference.
I couldn't find any solutions; however, I was able to find near misses such as 37, 397, 3997, or generally, a 3 followed by a string of 9s followed by a 7, which all differ by just one.
I am wondering if there actually are any solutions to this problem, and if not, how you would go about proving that.
Thank you!
number-theory linear-algebra
asked Jul 23 at 4:30
guy600
253
asked Jul 23 at 4:30
guy600
253
asked Jul 23 at 4:30
guy600
253
asked Jul 23 at 4:30
guy600
253
253
migrated from mathoverflow.net Jul 23 at 5:39
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Jul 23 at 5:39
This question came from our site for professional mathematicians.
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
We may represent integer $ n>0 $ as $ n = bcdot 10^d+Mcdot 10 + a $
(so that $ a>0 $ or $ M>0 $ or else we would have $ a=M=b=0) $
so that
$$ 2cdot(a!cdot! 10^d + M!cdot! 10 + b)
= b!cdot 10^d+M!cdot 10 + a $$
where $ a b $ are decimal digits (i.e. $ 0ldots 9), $ and $ M $ is
a non-negative integer such that $ M<10^d-1.$ Equivalently,
$$ (2!cdot! 10^d-1)!cdot a + 10!cdot!M
, =, (10^d-2)!cdot! b $$
or
$$ b, =, 2!cdot!a + frac10!cdot!M + 3!cdot!a10^d-2 $$
Thus,
$$ 0 < b-2cdot a = frac10!cdot!M + 3!cdot!a10^d-2
le 1 + frac3cdot a-810^d-2 $$
Also if we had $ dge 2 $ then this very last fraction has an
absolute value $ < 1 $ so that
$$ frac10!cdot!M + 3!cdot!a10^d-2 = 1 $$
while, knowing that $ a<5 $ (of course!),
$$ 10!cdot!M + 3!cdot!a, notequiv
, 10^d-2quad mod 10 $$
This contradiction shows that $ dle 1. $ Thus there are
only 2-digit solutions (if any).
In particular, $ M=0. $ This simplifies the equation:
$$ 19cdot a = 8cdot b $$
However, $19$ does not divide $ 8cdot b. $ Thus, after all,
$qquadqquad$ THERE ARE NO SOLUTIONS.
answered Jul 23 at 5:50
Wlod AA
3586
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We may represent integer $ n>0 $ as $ n = bcdot 10^d+Mcdot 10 + a $
(so that $ a>0 $ or $ M>0 $ or else we would have $ a=M=b=0) $
so that
$$ 2cdot(a!cdot! 10^d + M!cdot! 10 + b)
= b!cdot 10^d+M!cdot 10 + a $$
where $ a b $ are decimal digits (i.e. $ 0ldots 9), $ and $ M $ is
a non-negative integer such that $ M<10^d-1.$ Equivalently,
$$ (2!cdot! 10^d-1)!cdot a + 10!cdot!M
, =, (10^d-2)!cdot! b $$
or
$$ b, =, 2!cdot!a + frac10!cdot!M + 3!cdot!a10^d-2 $$
Thus,
$$ 0 < b-2cdot a = frac10!cdot!M + 3!cdot!a10^d-2
le 1 + frac3cdot a-810^d-2 $$
Also if we had $ dge 2 $ then this very last fraction has an
absolute value $ < 1 $ so that
$$ frac10!cdot!M + 3!cdot!a10^d-2 = 1 $$
while, knowing that $ a<5 $ (of course!),
$$ 10!cdot!M + 3!cdot!a, notequiv
, 10^d-2quad mod 10 $$
This contradiction shows that $ dle 1. $ Thus there are
only 2-digit solutions (if any).
In particular, $ M=0. $ This simplifies the equation:
$$ 19cdot a = 8cdot b $$
However, $19$ does not divide $ 8cdot b. $ Thus, after all,
$qquadqquad$ THERE ARE NO SOLUTIONS.
answered Jul 23 at 5:50
Wlod AA
3586
add a comment |Â
up vote
1
down vote
accepted
We may represent integer $ n>0 $ as $ n = bcdot 10^d+Mcdot 10 + a $
(so that $ a>0 $ or $ M>0 $ or else we would have $ a=M=b=0) $
so that
$$ 2cdot(a!cdot! 10^d + M!cdot! 10 + b)
= b!cdot 10^d+M!cdot 10 + a $$
where $ a b $ are decimal digits (i.e. $ 0ldots 9), $ and $ M $ is
a non-negative integer such that $ M<10^d-1.$ Equivalently,
$$ (2!cdot! 10^d-1)!cdot a + 10!cdot!M
, =, (10^d-2)!cdot! b $$
or
$$ b, =, 2!cdot!a + frac10!cdot!M + 3!cdot!a10^d-2 $$
Thus,
$$ 0 < b-2cdot a = frac10!cdot!M + 3!cdot!a10^d-2
le 1 + frac3cdot a-810^d-2 $$
Also if we had $ dge 2 $ then this very last fraction has an
absolute value $ < 1 $ so that
$$ frac10!cdot!M + 3!cdot!a10^d-2 = 1 $$
while, knowing that $ a<5 $ (of course!),
$$ 10!cdot!M + 3!cdot!a, notequiv
, 10^d-2quad mod 10 $$
This contradiction shows that $ dle 1. $ Thus there are
only 2-digit solutions (if any).
In particular, $ M=0. $ This simplifies the equation:
$$ 19cdot a = 8cdot b $$
However, $19$ does not divide $ 8cdot b. $ Thus, after all,
$qquadqquad$ THERE ARE NO SOLUTIONS.
answered Jul 23 at 5:50
Wlod AA
3586
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We may represent integer $ n>0 $ as $ n = bcdot 10^d+Mcdot 10 + a $
(so that $ a>0 $ or $ M>0 $ or else we would have $ a=M=b=0) $
so that
$$ 2cdot(a!cdot! 10^d + M!cdot! 10 + b)
= b!cdot 10^d+M!cdot 10 + a $$
where $ a b $ are decimal digits (i.e. $ 0ldots 9), $ and $ M $ is
a non-negative integer such that $ M<10^d-1.$ Equivalently,
$$ (2!cdot! 10^d-1)!cdot a + 10!cdot!M
, =, (10^d-2)!cdot! b $$
or
$$ b, =, 2!cdot!a + frac10!cdot!M + 3!cdot!a10^d-2 $$
Thus,
$$ 0 < b-2cdot a = frac10!cdot!M + 3!cdot!a10^d-2
le 1 + frac3cdot a-810^d-2 $$
Also if we had $ dge 2 $ then this very last fraction has an
absolute value $ < 1 $ so that
$$ frac10!cdot!M + 3!cdot!a10^d-2 = 1 $$
while, knowing that $ a<5 $ (of course!),
$$ 10!cdot!M + 3!cdot!a, notequiv
, 10^d-2quad mod 10 $$
This contradiction shows that $ dle 1. $ Thus there are
only 2-digit solutions (if any).
In particular, $ M=0. $ This simplifies the equation:
$$ 19cdot a = 8cdot b $$
However, $19$ does not divide $ 8cdot b. $ Thus, after all,
$qquadqquad$ THERE ARE NO SOLUTIONS.
answered Jul 23 at 5:50
Wlod AA
3586
We may represent integer $ n>0 $ as $ n = bcdot 10^d+Mcdot 10 + a $
(so that $ a>0 $ or $ M>0 $ or else we would have $ a=M=b=0) $
so that
$$ 2cdot(a!cdot! 10^d + M!cdot! 10 + b)
= b!cdot 10^d+M!cdot 10 + a $$
where $ a b $ are decimal digits (i.e. $ 0ldots 9), $ and $ M $ is
a non-negative integer such that $ M<10^d-1.$ Equivalently,
$$ (2!cdot! 10^d-1)!cdot a + 10!cdot!M
, =, (10^d-2)!cdot! b $$
or
$$ b, =, 2!cdot!a + frac10!cdot!M + 3!cdot!a10^d-2 $$
Thus,
$$ 0 < b-2cdot a = frac10!cdot!M + 3!cdot!a10^d-2
le 1 + frac3cdot a-810^d-2 $$
Also if we had $ dge 2 $ then this very last fraction has an
absolute value $ < 1 $ so that
$$ frac10!cdot!M + 3!cdot!a10^d-2 = 1 $$
while, knowing that $ a<5 $ (of course!),
$$ 10!cdot!M + 3!cdot!a, notequiv
, 10^d-2quad mod 10 $$
This contradiction shows that $ dle 1. $ Thus there are
only 2-digit solutions (if any).
In particular, $ M=0. $ This simplifies the equation:
$$ 19cdot a = 8cdot b $$
However, $19$ does not divide $ 8cdot b. $ Thus, after all,
$qquadqquad$ THERE ARE NO SOLUTIONS.
answered Jul 23 at 5:50
Wlod AA
3586
edited Jul 24 at 7:03
answered Jul 23 at 5:50
Wlod AA
3586
answered Jul 23 at 5:50
Wlod AA
3586
answered Jul 23 at 5:50
Wlod AA
3586
3586
add a comment |Â
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