The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing? [closed]
$begingroup$
The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing?
I came across this question in a math competition and I am looking for how to solve this question without working it out manually. Thanks.
contest-math
edited Jun 3 '14 at 19:43
Relure
2,1821035
asked Jun 3 '14 at 16:01
snivysteelsnivysteel
648620
$endgroup$
closed as off-topic by TheSimpliFire, Holo, amWhy, Xander Henderson, Gibbs Oct 21 '18 at 20:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, amWhy, Xander Henderson, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing?
I came across this question in a math competition and I am looking for how to solve this question without working it out manually. Thanks.
contest-math
edited Jun 3 '14 at 19:43
Relure
2,1821035
asked Jun 3 '14 at 16:01
snivysteelsnivysteel
648620
$endgroup$
closed as off-topic by TheSimpliFire, Holo, amWhy, Xander Henderson, Gibbs Oct 21 '18 at 20:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, amWhy, Xander Henderson, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
The accepted answer only works for numbers in which every digit is distinct. Is that supposed to be implied by "exactly 9 distinct digits"? I interpret that as also allowing a 10-digit number with one duplicate.
$endgroup$
– Brilliand
Jun 3 '14 at 18:33
$begingroup$
"Has exactly 9 digits, all of which are distinct" might be more accurate.
$endgroup$
– Dan
Jun 3 '14 at 19:32
2
$begingroup$
@Brilliand: if we start with $2^{10}=1024$, it is not too hard to see that $2^{29}approx500,000,000$. So, $2^{29}$ has $9$ digits.
$endgroup$
– robjohn♦
Jun 3 '14 at 19:54
$begingroup$
I see how we know it has 9digits, $29log 2=8.73$ but how does one know the digits are distinct ?
$endgroup$
– Rene Schipperus
Jun 3 '14 at 22:38
$begingroup$
Just asking: Is it (the competition) SMO?
$endgroup$
– user148697
Jun 27 '14 at 13:42
|
show 1 more comment
$begingroup$
The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing?
I came across this question in a math competition and I am looking for how to solve this question without working it out manually. Thanks.
contest-math
edited Jun 3 '14 at 19:43
Relure
2,1821035
asked Jun 3 '14 at 16:01
snivysteelsnivysteel
648620
$endgroup$
The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing?
I came across this question in a math competition and I am looking for how to solve this question without working it out manually. Thanks.
contest-math
contest-math
edited Jun 3 '14 at 19:43
Relure
2,1821035
asked Jun 3 '14 at 16:01
snivysteelsnivysteel
648620
edited Jun 3 '14 at 19:43
Relure
2,1821035
asked Jun 3 '14 at 16:01
snivysteelsnivysteel
648620
edited Jun 3 '14 at 19:43
Relure
2,1821035
edited Jun 3 '14 at 19:43
Relure
2,1821035
edited Jun 3 '14 at 19:43
Relure
2,1821035
2,1821035
asked Jun 3 '14 at 16:01
snivysteelsnivysteel
648620
asked Jun 3 '14 at 16:01
snivysteelsnivysteel
648620
asked Jun 3 '14 at 16:01
snivysteelsnivysteel
648620
648620
closed as off-topic by TheSimpliFire, Holo, amWhy, Xander Henderson, Gibbs Oct 21 '18 at 20:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, amWhy, Xander Henderson, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by TheSimpliFire, Holo, amWhy, Xander Henderson, Gibbs Oct 21 '18 at 20:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, amWhy, Xander Henderson, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
The accepted answer only works for numbers in which every digit is distinct. Is that supposed to be implied by "exactly 9 distinct digits"? I interpret that as also allowing a 10-digit number with one duplicate.
$endgroup$
– Brilliand
Jun 3 '14 at 18:33
$begingroup$
"Has exactly 9 digits, all of which are distinct" might be more accurate.
$endgroup$
– Dan
Jun 3 '14 at 19:32
2
$begingroup$
@Brilliand: if we start with $2^{10}=1024$, it is not too hard to see that $2^{29}approx500,000,000$. So, $2^{29}$ has $9$ digits.
$endgroup$
– robjohn♦
Jun 3 '14 at 19:54
$begingroup$
I see how we know it has 9digits, $29log 2=8.73$ but how does one know the digits are distinct ?
$endgroup$
– Rene Schipperus
Jun 3 '14 at 22:38
$begingroup$
Just asking: Is it (the competition) SMO?
$endgroup$
– user148697
Jun 27 '14 at 13:42
|
show 1 more comment
4
$begingroup$
The accepted answer only works for numbers in which every digit is distinct. Is that supposed to be implied by "exactly 9 distinct digits"? I interpret that as also allowing a 10-digit number with one duplicate.
$endgroup$
– Brilliand
Jun 3 '14 at 18:33
$begingroup$
"Has exactly 9 digits, all of which are distinct" might be more accurate.
$endgroup$
– Dan
Jun 3 '14 at 19:32
2
$begingroup$
@Brilliand: if we start with $2^{10}=1024$, it is not too hard to see that $2^{29}approx500,000,000$. So, $2^{29}$ has $9$ digits.
$endgroup$
– robjohn♦
Jun 3 '14 at 19:54
$begingroup$
I see how we know it has 9digits, $29log 2=8.73$ but how does one know the digits are distinct ?
$endgroup$
– Rene Schipperus
Jun 3 '14 at 22:38
$begingroup$
Just asking: Is it (the competition) SMO?
$endgroup$
– user148697
Jun 27 '14 at 13:42
4
4
$begingroup$
The accepted answer only works for numbers in which every digit is distinct. Is that supposed to be implied by "exactly 9 distinct digits"? I interpret that as also allowing a 10-digit number with one duplicate.
$endgroup$
– Brilliand
Jun 3 '14 at 18:33
$begingroup$
The accepted answer only works for numbers in which every digit is distinct. Is that supposed to be implied by "exactly 9 distinct digits"? I interpret that as also allowing a 10-digit number with one duplicate.
$endgroup$
– Brilliand
Jun 3 '14 at 18:33
$begingroup$
"Has exactly 9 digits, all of which are distinct" might be more accurate.
$endgroup$
– Dan
Jun 3 '14 at 19:32
$begingroup$
"Has exactly 9 digits, all of which are distinct" might be more accurate.
$endgroup$
– Dan
Jun 3 '14 at 19:32
2
2
$begingroup$
@Brilliand: if we start with $2^{10}=1024$, it is not too hard to see that $2^{29}approx500,000,000$. So, $2^{29}$ has $9$ digits.
$endgroup$
– robjohn♦
Jun 3 '14 at 19:54
$begingroup$
@Brilliand: if we start with $2^{10}=1024$, it is not too hard to see that $2^{29}approx500,000,000$. So, $2^{29}$ has $9$ digits.
$endgroup$
– robjohn♦
Jun 3 '14 at 19:54
$begingroup$
I see how we know it has 9digits, $29log 2=8.73$ but how does one know the digits are distinct ?
$endgroup$
– Rene Schipperus
Jun 3 '14 at 22:38
$begingroup$
I see how we know it has 9digits, $29log 2=8.73$ but how does one know the digits are distinct ?
$endgroup$
– Rene Schipperus
Jun 3 '14 at 22:38
$begingroup$
Just asking: Is it (the competition) SMO?
$endgroup$
– user148697
Jun 27 '14 at 13:42
$begingroup$
Just asking: Is it (the competition) SMO?
$endgroup$
– user148697
Jun 27 '14 at 13:42
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$$sum k_n 10^n equiv sum k_n mod 9$$
the sum of all the digits is $frac{9(9+1)}{2}equiv 0 mod 9$ so the sum of all but one $x$ is $equiv -x mod 9$
Now $$2^{29}equiv -4 mod 9$$ so $4$ is the missing digit.
edited Jun 3 '14 at 16:41
Hakim
9,19953157
answered Jun 3 '14 at 16:38
Rene SchipperusRene Schipperus
32.3k11960
$endgroup$
$begingroup$
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} pmod 9$. Any hint will also suffice! Thanks for the answer, by the way!
$endgroup$
– puru
Jun 28 '14 at 2:51
$begingroup$
You find the power of two mod 9, that is $2^nequiv 1pmod 9$ I forget what it is but then you divide it into 29.
$endgroup$
– Rene Schipperus
Jun 28 '14 at 2:58
2
$begingroup$
Thanks, I got it. In fact, $2^{29}=2^{27} times 4=(9-1)^9 times 4$ which implies that $2^{29} pmod 9= -4$
$endgroup$
– puru
Jun 28 '14 at 9:29
add a comment |
$begingroup$
A hint: Think about the remainder modulo $9$.
answered Jun 3 '14 at 16:26
Christian BlatterChristian Blatter
174k8115327
$endgroup$
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$endgroup$
– Rene Schipperus
Jun 3 '14 at 16:32
add a comment |
$begingroup$
$2^{29}$ is not that big. You can just compute it. A fast launch point is to know that $2^{10}=1024$. So you just need to multiply $1024cdot1024cdot512$, which can be done by hand quickly in a competition.
edited Jun 3 '14 at 17:57
nbarto
14.1k32682
answered Jun 3 '14 at 16:06
alex.jordanalex.jordan
39.3k560121
$endgroup$
7
$begingroup$
It's about creative solution not answering it
$endgroup$
– Karo
Jun 3 '14 at 16:08
19
$begingroup$
Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work?
$endgroup$
– Jorge Fernández Hidalgo
Jun 3 '14 at 16:11
$begingroup$
whats should we do? don't think on other methods to solve it?
$endgroup$
– Karo
Jun 3 '14 at 16:22
$begingroup$
@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever?
$endgroup$
– alex.jordan
Jun 3 '14 at 16:38
2
$begingroup$
It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this.
$endgroup$
– Karo
Jun 3 '14 at 17:36
add a comment |
$begingroup$
Let the number $N$ be represented by $9$ digits $d_i$ for $i=0,dots,8$ so that
$$N = sum_{i=0}^{8} d_i 10^i$$
We first notice that
$$sum_{i=0}^{8} d_i 10^i equiv sum_i d_i pmod{9}$$
Since only one digit is missing, this sum must be between 36 (with 9 missing) and 45 (with 0 missing). There are two cases to consider. Either $sum d_i equiv 0 pmod{9}$ or it is not. In the first case, either 0 or 9 is the missing digit, since those are the only missing values for which $sum_i d_i equiv 0 pmod{9}$, making the sum either 36 or 45. We can determine which is the value by looking at the sum $pmod{8} = N pmod{8}$: if this value is 4 then the sum is 36 and 9 is the missing digit. If $N equiv 5 pmod{8}$, then the sum is 45 and 0 is the missing digit.
In the second case, we note that the sum of all but one digit $x$ is congruent to $-x pmod{9}$. Solving this congruence for $x$ gives the missing digit.
This addresses the case of an arbitrary $N$ for which $N equiv 0 pmod{9}$ which the accepted answer does not handle correctly. Since I'm not yet allowed to comment or improve other answers by censors, I just constructed a new, complete answer. Vote it up so in the future I can do this the right way.
answered Jun 3 '14 at 19:39
jspencerjspencer
255
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1
$begingroup$
This looks eerily similar to Rene Schipperus' answer.
$endgroup$
– robjohn♦
Jun 3 '14 at 20:00
$begingroup$
It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$...
$endgroup$
– user21820
Jun 4 '14 at 11:58
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$$sum k_n 10^n equiv sum k_n mod 9$$
the sum of all the digits is $frac{9(9+1)}{2}equiv 0 mod 9$ so the sum of all but one $x$ is $equiv -x mod 9$
Now $$2^{29}equiv -4 mod 9$$ so $4$ is the missing digit.
edited Jun 3 '14 at 16:41
Hakim
9,19953157
answered Jun 3 '14 at 16:38
Rene SchipperusRene Schipperus
32.3k11960
$endgroup$
$begingroup$
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} pmod 9$. Any hint will also suffice! Thanks for the answer, by the way!
$endgroup$
– puru
Jun 28 '14 at 2:51
$begingroup$
You find the power of two mod 9, that is $2^nequiv 1pmod 9$ I forget what it is but then you divide it into 29.
$endgroup$
– Rene Schipperus
Jun 28 '14 at 2:58
2
$begingroup$
Thanks, I got it. In fact, $2^{29}=2^{27} times 4=(9-1)^9 times 4$ which implies that $2^{29} pmod 9= -4$
$endgroup$
– puru
Jun 28 '14 at 9:29
add a comment |
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$$sum k_n 10^n equiv sum k_n mod 9$$
the sum of all the digits is $frac{9(9+1)}{2}equiv 0 mod 9$ so the sum of all but one $x$ is $equiv -x mod 9$
Now $$2^{29}equiv -4 mod 9$$ so $4$ is the missing digit.
edited Jun 3 '14 at 16:41
Hakim
9,19953157
answered Jun 3 '14 at 16:38
Rene SchipperusRene Schipperus
32.3k11960
$endgroup$
$begingroup$
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} pmod 9$. Any hint will also suffice! Thanks for the answer, by the way!
$endgroup$
– puru
Jun 28 '14 at 2:51
$begingroup$
You find the power of two mod 9, that is $2^nequiv 1pmod 9$ I forget what it is but then you divide it into 29.
$endgroup$
– Rene Schipperus
Jun 28 '14 at 2:58
2
$begingroup$
Thanks, I got it. In fact, $2^{29}=2^{27} times 4=(9-1)^9 times 4$ which implies that $2^{29} pmod 9= -4$
$endgroup$
– puru
Jun 28 '14 at 9:29
add a comment |
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$$sum k_n 10^n equiv sum k_n mod 9$$
the sum of all the digits is $frac{9(9+1)}{2}equiv 0 mod 9$ so the sum of all but one $x$ is $equiv -x mod 9$
Now $$2^{29}equiv -4 mod 9$$ so $4$ is the missing digit.
edited Jun 3 '14 at 16:41
Hakim
9,19953157
answered Jun 3 '14 at 16:38
Rene SchipperusRene Schipperus
32.3k11960
$endgroup$
Oh its so easy, now that we follow the hint (thanks !)
$$sum k_n 10^n equiv sum k_n mod 9$$
the sum of all the digits is $frac{9(9+1)}{2}equiv 0 mod 9$ so the sum of all but one $x$ is $equiv -x mod 9$
Now $$2^{29}equiv -4 mod 9$$ so $4$ is the missing digit.
edited Jun 3 '14 at 16:41
Hakim
9,19953157
answered Jun 3 '14 at 16:38
Rene SchipperusRene Schipperus
32.3k11960
edited Jun 3 '14 at 16:41
Hakim
9,19953157
edited Jun 3 '14 at 16:41
Hakim
9,19953157
edited Jun 3 '14 at 16:41
Hakim
9,19953157
9,19953157
answered Jun 3 '14 at 16:38
Rene SchipperusRene Schipperus
32.3k11960
answered Jun 3 '14 at 16:38
Rene SchipperusRene Schipperus
32.3k11960
answered Jun 3 '14 at 16:38
Rene SchipperusRene Schipperus
32.3k11960
32.3k11960
$begingroup$
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} pmod 9$. Any hint will also suffice! Thanks for the answer, by the way!
$endgroup$
– puru
Jun 28 '14 at 2:51
$begingroup$
You find the power of two mod 9, that is $2^nequiv 1pmod 9$ I forget what it is but then you divide it into 29.
$endgroup$
– Rene Schipperus
Jun 28 '14 at 2:58
2
$begingroup$
Thanks, I got it. In fact, $2^{29}=2^{27} times 4=(9-1)^9 times 4$ which implies that $2^{29} pmod 9= -4$
$endgroup$
– puru
Jun 28 '14 at 9:29
add a comment |
$begingroup$
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} pmod 9$. Any hint will also suffice! Thanks for the answer, by the way!
$endgroup$
– puru
Jun 28 '14 at 2:51
$begingroup$
You find the power of two mod 9, that is $2^nequiv 1pmod 9$ I forget what it is but then you divide it into 29.
$endgroup$
– Rene Schipperus
Jun 28 '14 at 2:58
2
$begingroup$
Thanks, I got it. In fact, $2^{29}=2^{27} times 4=(9-1)^9 times 4$ which implies that $2^{29} pmod 9= -4$
$endgroup$
– puru
Jun 28 '14 at 9:29
$begingroup$
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} pmod 9$. Any hint will also suffice! Thanks for the answer, by the way!
$endgroup$
– puru
Jun 28 '14 at 2:51
$begingroup$
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} pmod 9$. Any hint will also suffice! Thanks for the answer, by the way!
$endgroup$
– puru
Jun 28 '14 at 2:51
$begingroup$
You find the power of two mod 9, that is $2^nequiv 1pmod 9$ I forget what it is but then you divide it into 29.
$endgroup$
– Rene Schipperus
Jun 28 '14 at 2:58
$begingroup$
You find the power of two mod 9, that is $2^nequiv 1pmod 9$ I forget what it is but then you divide it into 29.
$endgroup$
– Rene Schipperus
Jun 28 '14 at 2:58
2
2
$begingroup$
Thanks, I got it. In fact, $2^{29}=2^{27} times 4=(9-1)^9 times 4$ which implies that $2^{29} pmod 9= -4$
$endgroup$
– puru
Jun 28 '14 at 9:29
$begingroup$
Thanks, I got it. In fact, $2^{29}=2^{27} times 4=(9-1)^9 times 4$ which implies that $2^{29} pmod 9= -4$
$endgroup$
– puru
Jun 28 '14 at 9:29
add a comment |
$begingroup$
A hint: Think about the remainder modulo $9$.
answered Jun 3 '14 at 16:26
Christian BlatterChristian Blatter
174k8115327
$endgroup$
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$endgroup$
– Rene Schipperus
Jun 3 '14 at 16:32
add a comment |
$begingroup$
A hint: Think about the remainder modulo $9$.
answered Jun 3 '14 at 16:26
Christian BlatterChristian Blatter
174k8115327
$endgroup$
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$endgroup$
– Rene Schipperus
Jun 3 '14 at 16:32
add a comment |
$begingroup$
A hint: Think about the remainder modulo $9$.
answered Jun 3 '14 at 16:26
Christian BlatterChristian Blatter
174k8115327
$endgroup$
A hint: Think about the remainder modulo $9$.
answered Jun 3 '14 at 16:26
Christian BlatterChristian Blatter
174k8115327
answered Jun 3 '14 at 16:26
Christian BlatterChristian Blatter
174k8115327
answered Jun 3 '14 at 16:26
Christian BlatterChristian Blatter
174k8115327
answered Jun 3 '14 at 16:26
Christian BlatterChristian Blatter
174k8115327
174k8115327
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$endgroup$
– Rene Schipperus
Jun 3 '14 at 16:32
add a comment |
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$endgroup$
– Rene Schipperus
Jun 3 '14 at 16:32
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$endgroup$
– Rene Schipperus
Jun 3 '14 at 16:32
$begingroup$
Oh its so easy, now that we follow the hint (thanks !)
$endgroup$
– Rene Schipperus
Jun 3 '14 at 16:32
add a comment |
$begingroup$
$2^{29}$ is not that big. You can just compute it. A fast launch point is to know that $2^{10}=1024$. So you just need to multiply $1024cdot1024cdot512$, which can be done by hand quickly in a competition.
edited Jun 3 '14 at 17:57
nbarto
14.1k32682
answered Jun 3 '14 at 16:06
alex.jordanalex.jordan
39.3k560121
$endgroup$
7
$begingroup$
It's about creative solution not answering it
$endgroup$
– Karo
Jun 3 '14 at 16:08
19
$begingroup$
Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work?
$endgroup$
– Jorge Fernández Hidalgo
Jun 3 '14 at 16:11
$begingroup$
whats should we do? don't think on other methods to solve it?
$endgroup$
– Karo
Jun 3 '14 at 16:22
$begingroup$
@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever?
$endgroup$
– alex.jordan
Jun 3 '14 at 16:38
2
$begingroup$
It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this.
$endgroup$
– Karo
Jun 3 '14 at 17:36
add a comment |
$begingroup$
$2^{29}$ is not that big. You can just compute it. A fast launch point is to know that $2^{10}=1024$. So you just need to multiply $1024cdot1024cdot512$, which can be done by hand quickly in a competition.
edited Jun 3 '14 at 17:57
nbarto
14.1k32682
answered Jun 3 '14 at 16:06
alex.jordanalex.jordan
39.3k560121
$endgroup$
7
$begingroup$
It's about creative solution not answering it
$endgroup$
– Karo
Jun 3 '14 at 16:08
19
$begingroup$
Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work?
$endgroup$
– Jorge Fernández Hidalgo
Jun 3 '14 at 16:11
$begingroup$
whats should we do? don't think on other methods to solve it?
$endgroup$
– Karo
Jun 3 '14 at 16:22
$begingroup$
@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever?
$endgroup$
– alex.jordan
Jun 3 '14 at 16:38
2
$begingroup$
It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this.
$endgroup$
– Karo
Jun 3 '14 at 17:36
add a comment |
$begingroup$
$2^{29}$ is not that big. You can just compute it. A fast launch point is to know that $2^{10}=1024$. So you just need to multiply $1024cdot1024cdot512$, which can be done by hand quickly in a competition.
edited Jun 3 '14 at 17:57
nbarto
14.1k32682
answered Jun 3 '14 at 16:06
alex.jordanalex.jordan
39.3k560121
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$2^{29}$ is not that big. You can just compute it. A fast launch point is to know that $2^{10}=1024$. So you just need to multiply $1024cdot1024cdot512$, which can be done by hand quickly in a competition.
edited Jun 3 '14 at 17:57
nbarto
14.1k32682
answered Jun 3 '14 at 16:06
alex.jordanalex.jordan
39.3k560121
edited Jun 3 '14 at 17:57
nbarto
14.1k32682
edited Jun 3 '14 at 17:57
nbarto
14.1k32682
edited Jun 3 '14 at 17:57
nbarto
14.1k32682
14.1k32682
answered Jun 3 '14 at 16:06
alex.jordanalex.jordan
39.3k560121
answered Jun 3 '14 at 16:06
alex.jordanalex.jordan
39.3k560121
answered Jun 3 '14 at 16:06
alex.jordanalex.jordan
39.3k560121
39.3k560121
7
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It's about creative solution not answering it
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– Karo
Jun 3 '14 at 16:08
19
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Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work?
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– Jorge Fernández Hidalgo
Jun 3 '14 at 16:11
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whats should we do? don't think on other methods to solve it?
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– Karo
Jun 3 '14 at 16:22
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@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever?
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– alex.jordan
Jun 3 '14 at 16:38
2
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It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this.
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– Karo
Jun 3 '14 at 17:36
add a comment |
7
$begingroup$
It's about creative solution not answering it
$endgroup$
– Karo
Jun 3 '14 at 16:08
19
$begingroup$
Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work?
$endgroup$
– Jorge Fernández Hidalgo
Jun 3 '14 at 16:11
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whats should we do? don't think on other methods to solve it?
$endgroup$
– Karo
Jun 3 '14 at 16:22
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@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever?
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– alex.jordan
Jun 3 '14 at 16:38
2
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It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this.
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– Karo
Jun 3 '14 at 17:36
7
7
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It's about creative solution not answering it
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– Karo
Jun 3 '14 at 16:08
$begingroup$
It's about creative solution not answering it
$endgroup$
– Karo
Jun 3 '14 at 16:08
19
19
$begingroup$
Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work?
$endgroup$
– Jorge Fernández Hidalgo
Jun 3 '14 at 16:11
$begingroup$
Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work?
$endgroup$
– Jorge Fernández Hidalgo
Jun 3 '14 at 16:11
$begingroup$
whats should we do? don't think on other methods to solve it?
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– Karo
Jun 3 '14 at 16:22
$begingroup$
whats should we do? don't think on other methods to solve it?
$endgroup$
– Karo
Jun 3 '14 at 16:22
$begingroup$
@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever?
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– alex.jordan
Jun 3 '14 at 16:38
$begingroup$
@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever?
$endgroup$
– alex.jordan
Jun 3 '14 at 16:38
2
2
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It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this.
$endgroup$
– Karo
Jun 3 '14 at 17:36
$begingroup$
It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this.
$endgroup$
– Karo
Jun 3 '14 at 17:36
add a comment |
$begingroup$
Let the number $N$ be represented by $9$ digits $d_i$ for $i=0,dots,8$ so that
$$N = sum_{i=0}^{8} d_i 10^i$$
We first notice that
$$sum_{i=0}^{8} d_i 10^i equiv sum_i d_i pmod{9}$$
Since only one digit is missing, this sum must be between 36 (with 9 missing) and 45 (with 0 missing). There are two cases to consider. Either $sum d_i equiv 0 pmod{9}$ or it is not. In the first case, either 0 or 9 is the missing digit, since those are the only missing values for which $sum_i d_i equiv 0 pmod{9}$, making the sum either 36 or 45. We can determine which is the value by looking at the sum $pmod{8} = N pmod{8}$: if this value is 4 then the sum is 36 and 9 is the missing digit. If $N equiv 5 pmod{8}$, then the sum is 45 and 0 is the missing digit.
In the second case, we note that the sum of all but one digit $x$ is congruent to $-x pmod{9}$. Solving this congruence for $x$ gives the missing digit.
This addresses the case of an arbitrary $N$ for which $N equiv 0 pmod{9}$ which the accepted answer does not handle correctly. Since I'm not yet allowed to comment or improve other answers by censors, I just constructed a new, complete answer. Vote it up so in the future I can do this the right way.
answered Jun 3 '14 at 19:39
jspencerjspencer
255
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1
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This looks eerily similar to Rene Schipperus' answer.
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– robjohn♦
Jun 3 '14 at 20:00
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It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$...
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– user21820
Jun 4 '14 at 11:58
add a comment |
$begingroup$
Let the number $N$ be represented by $9$ digits $d_i$ for $i=0,dots,8$ so that
$$N = sum_{i=0}^{8} d_i 10^i$$
We first notice that
$$sum_{i=0}^{8} d_i 10^i equiv sum_i d_i pmod{9}$$
Since only one digit is missing, this sum must be between 36 (with 9 missing) and 45 (with 0 missing). There are two cases to consider. Either $sum d_i equiv 0 pmod{9}$ or it is not. In the first case, either 0 or 9 is the missing digit, since those are the only missing values for which $sum_i d_i equiv 0 pmod{9}$, making the sum either 36 or 45. We can determine which is the value by looking at the sum $pmod{8} = N pmod{8}$: if this value is 4 then the sum is 36 and 9 is the missing digit. If $N equiv 5 pmod{8}$, then the sum is 45 and 0 is the missing digit.
In the second case, we note that the sum of all but one digit $x$ is congruent to $-x pmod{9}$. Solving this congruence for $x$ gives the missing digit.
This addresses the case of an arbitrary $N$ for which $N equiv 0 pmod{9}$ which the accepted answer does not handle correctly. Since I'm not yet allowed to comment or improve other answers by censors, I just constructed a new, complete answer. Vote it up so in the future I can do this the right way.
answered Jun 3 '14 at 19:39
jspencerjspencer
255
$endgroup$
1
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This looks eerily similar to Rene Schipperus' answer.
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– robjohn♦
Jun 3 '14 at 20:00
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It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$...
$endgroup$
– user21820
Jun 4 '14 at 11:58
add a comment |
$begingroup$
Let the number $N$ be represented by $9$ digits $d_i$ for $i=0,dots,8$ so that
$$N = sum_{i=0}^{8} d_i 10^i$$
We first notice that
$$sum_{i=0}^{8} d_i 10^i equiv sum_i d_i pmod{9}$$
Since only one digit is missing, this sum must be between 36 (with 9 missing) and 45 (with 0 missing). There are two cases to consider. Either $sum d_i equiv 0 pmod{9}$ or it is not. In the first case, either 0 or 9 is the missing digit, since those are the only missing values for which $sum_i d_i equiv 0 pmod{9}$, making the sum either 36 or 45. We can determine which is the value by looking at the sum $pmod{8} = N pmod{8}$: if this value is 4 then the sum is 36 and 9 is the missing digit. If $N equiv 5 pmod{8}$, then the sum is 45 and 0 is the missing digit.
In the second case, we note that the sum of all but one digit $x$ is congruent to $-x pmod{9}$. Solving this congruence for $x$ gives the missing digit.
This addresses the case of an arbitrary $N$ for which $N equiv 0 pmod{9}$ which the accepted answer does not handle correctly. Since I'm not yet allowed to comment or improve other answers by censors, I just constructed a new, complete answer. Vote it up so in the future I can do this the right way.
answered Jun 3 '14 at 19:39
jspencerjspencer
255
$endgroup$
Let the number $N$ be represented by $9$ digits $d_i$ for $i=0,dots,8$ so that
$$N = sum_{i=0}^{8} d_i 10^i$$
We first notice that
$$sum_{i=0}^{8} d_i 10^i equiv sum_i d_i pmod{9}$$
Since only one digit is missing, this sum must be between 36 (with 9 missing) and 45 (with 0 missing). There are two cases to consider. Either $sum d_i equiv 0 pmod{9}$ or it is not. In the first case, either 0 or 9 is the missing digit, since those are the only missing values for which $sum_i d_i equiv 0 pmod{9}$, making the sum either 36 or 45. We can determine which is the value by looking at the sum $pmod{8} = N pmod{8}$: if this value is 4 then the sum is 36 and 9 is the missing digit. If $N equiv 5 pmod{8}$, then the sum is 45 and 0 is the missing digit.
In the second case, we note that the sum of all but one digit $x$ is congruent to $-x pmod{9}$. Solving this congruence for $x$ gives the missing digit.
This addresses the case of an arbitrary $N$ for which $N equiv 0 pmod{9}$ which the accepted answer does not handle correctly. Since I'm not yet allowed to comment or improve other answers by censors, I just constructed a new, complete answer. Vote it up so in the future I can do this the right way.
answered Jun 3 '14 at 19:39
jspencerjspencer
255
edited Jun 3 '14 at 20:13
answered Jun 3 '14 at 19:39
jspencerjspencer
255
answered Jun 3 '14 at 19:39
jspencerjspencer
255
answered Jun 3 '14 at 19:39
jspencerjspencer
255
255
1
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This looks eerily similar to Rene Schipperus' answer.
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– robjohn♦
Jun 3 '14 at 20:00
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It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$...
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– user21820
Jun 4 '14 at 11:58
add a comment |
1
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This looks eerily similar to Rene Schipperus' answer.
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– robjohn♦
Jun 3 '14 at 20:00
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It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$...
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– user21820
Jun 4 '14 at 11:58
1
1
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This looks eerily similar to Rene Schipperus' answer.
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– robjohn♦
Jun 3 '14 at 20:00
$begingroup$
This looks eerily similar to Rene Schipperus' answer.
$endgroup$
– robjohn♦
Jun 3 '14 at 20:00
$begingroup$
It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$...
$endgroup$
– user21820
Jun 4 '14 at 11:58
$begingroup$
It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$...
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– user21820
Jun 4 '14 at 11:58
add a comment |
4
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The accepted answer only works for numbers in which every digit is distinct. Is that supposed to be implied by "exactly 9 distinct digits"? I interpret that as also allowing a 10-digit number with one duplicate.
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– Brilliand
Jun 3 '14 at 18:33
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"Has exactly 9 digits, all of which are distinct" might be more accurate.
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– Dan
Jun 3 '14 at 19:32
2
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@Brilliand: if we start with $2^{10}=1024$, it is not too hard to see that $2^{29}approx500,000,000$. So, $2^{29}$ has $9$ digits.
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– robjohn♦
Jun 3 '14 at 19:54
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I see how we know it has 9digits, $29log 2=8.73$ but how does one know the digits are distinct ?
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– Rene Schipperus
Jun 3 '14 at 22:38
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Just asking: Is it (the competition) SMO?
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– user148697
Jun 27 '14 at 13:42