Find a number such that $S(n)+P(n)=n$
Find a number $nin mathbb N$ such that $S(n)+P(n)=n$. Where $S(n)$ is sum of digits, and $P(n)$ is product of digits
I know that $9|n-S(n)$ ,since $n-S(n)=P(n)$ then $9|P(n)$ so at least one number must be nine or two digits $ldots33 ldots$ then I go with number $19,29,39,49,59,69,79,89,99$ for all this number $P(n)+S(n)=n$ but for number like 199 or 999 I just can not reach that number using formula, but I do not know how to prove it is not enough just to say that. And I do not know is there some other number can you help me?
elementary-number-theory
edited Nov 24 at 9:49
greedoid
37.7k114794
asked Nov 24 at 5:13
Marko Škorić
70310
add a comment |
Find a number $nin mathbb N$ such that $S(n)+P(n)=n$. Where $S(n)$ is sum of digits, and $P(n)$ is product of digits
I know that $9|n-S(n)$ ,since $n-S(n)=P(n)$ then $9|P(n)$ so at least one number must be nine or two digits $ldots33 ldots$ then I go with number $19,29,39,49,59,69,79,89,99$ for all this number $P(n)+S(n)=n$ but for number like 199 or 999 I just can not reach that number using formula, but I do not know how to prove it is not enough just to say that. And I do not know is there some other number can you help me?
elementary-number-theory
edited Nov 24 at 9:49
greedoid
37.7k114794
asked Nov 24 at 5:13
Marko Škorić
70310
add a comment |
Find a number $nin mathbb N$ such that $S(n)+P(n)=n$. Where $S(n)$ is sum of digits, and $P(n)$ is product of digits
I know that $9|n-S(n)$ ,since $n-S(n)=P(n)$ then $9|P(n)$ so at least one number must be nine or two digits $ldots33 ldots$ then I go with number $19,29,39,49,59,69,79,89,99$ for all this number $P(n)+S(n)=n$ but for number like 199 or 999 I just can not reach that number using formula, but I do not know how to prove it is not enough just to say that. And I do not know is there some other number can you help me?
elementary-number-theory
edited Nov 24 at 9:49
greedoid
37.7k114794
asked Nov 24 at 5:13
Marko Škorić
70310
Find a number $nin mathbb N$ such that $S(n)+P(n)=n$. Where $S(n)$ is sum of digits, and $P(n)$ is product of digits
I know that $9|n-S(n)$ ,since $n-S(n)=P(n)$ then $9|P(n)$ so at least one number must be nine or two digits $ldots33 ldots$ then I go with number $19,29,39,49,59,69,79,89,99$ for all this number $P(n)+S(n)=n$ but for number like 199 or 999 I just can not reach that number using formula, but I do not know how to prove it is not enough just to say that. And I do not know is there some other number can you help me?
elementary-number-theory
elementary-number-theory
edited Nov 24 at 9:49
greedoid
37.7k114794
asked Nov 24 at 5:13
Marko Škorić
70310
edited Nov 24 at 9:49
greedoid
37.7k114794
asked Nov 24 at 5:13
Marko Škorić
70310
edited Nov 24 at 9:49
greedoid
37.7k114794
edited Nov 24 at 9:49
greedoid
37.7k114794
edited Nov 24 at 9:49
greedoid
37.7k114794
37.7k114794
asked Nov 24 at 5:13
Marko Škorić
70310
asked Nov 24 at 5:13
Marko Škorić
70310
asked Nov 24 at 5:13
Marko Škorić
70310
70310
add a comment |
add a comment |
3 Answers
3
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A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:
$$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$
and
$$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$
In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction
$$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$
(Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)
answered Nov 24 at 12:04
Barry Cipra
59k653123
add a comment |
Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then
$$S(n)leq 10(k-1)+d$$
and
$$P(n)leq d9^{k-1},$$
so
$$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$
$$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$
$$10^{k-1}-9^{k-1}-1leq 10(k-1).$$
Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?
answered Nov 24 at 5:22
Carl Schildkraut
11.1k11441
add a comment |
Hint:
Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$
Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...
answered Nov 24 at 9:48
greedoid
37.7k114794
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
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votes
A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:
$$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$
and
$$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$
In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction
$$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$
(Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)
answered Nov 24 at 12:04
Barry Cipra
59k653123
add a comment |
A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:
$$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$
and
$$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$
In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction
$$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$
(Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)
answered Nov 24 at 12:04
Barry Cipra
59k653123
add a comment |
A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:
$$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$
and
$$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$
In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction
$$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$
(Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)
answered Nov 24 at 12:04
Barry Cipra
59k653123
A couple of examples should suffice to show that we cannot have $n=S(n)+P(n)$ if $n$ has more than two digits:
$$100a+10b+c=a+b+c+abcimplies c=9left({11over b}+{1over a} right)ge9cdot{11over10}gt9$$
and
$$1000a+100b+10c+d=a+b+c+d+abcdimplies d=9left({111over bc}+{11over ca}+{1over ab}right)ge9cdot{111over10cdot10}gt9$$
In general, if $n=a_k10^k+cdots+a_0$ with $kgt1$, then $n=S(n)+P(n)$ implies the contradiction
$$a_0ge9left((10^k-1)/9over a_{k-1}cdots a_1right)ge9left(1+10+cdots+10^{k-1}over10^{k-1} right)gt9$$
(Note, for $k=1$, we just get $a_0ge9$, which is not a contradiction.)
answered Nov 24 at 12:04
Barry Cipra
59k653123
answered Nov 24 at 12:04
Barry Cipra
59k653123
answered Nov 24 at 12:04
Barry Cipra
59k653123
answered Nov 24 at 12:04
Barry Cipra
59k653123
59k653123
add a comment |
add a comment |
Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then
$$S(n)leq 10(k-1)+d$$
and
$$P(n)leq d9^{k-1},$$
so
$$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$
$$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$
$$10^{k-1}-9^{k-1}-1leq 10(k-1).$$
Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?
answered Nov 24 at 5:22
Carl Schildkraut
11.1k11441
add a comment |
Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then
$$S(n)leq 10(k-1)+d$$
and
$$P(n)leq d9^{k-1},$$
so
$$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$
$$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$
$$10^{k-1}-9^{k-1}-1leq 10(k-1).$$
Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?
answered Nov 24 at 5:22
Carl Schildkraut
11.1k11441
add a comment |
Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then
$$S(n)leq 10(k-1)+d$$
and
$$P(n)leq d9^{k-1},$$
so
$$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$
$$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$
$$10^{k-1}-9^{k-1}-1leq 10(k-1).$$
Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?
answered Nov 24 at 5:22
Carl Schildkraut
11.1k11441
Hint: If a number $n$ is $k$ digits long and begins with the digit $d$, then
$$S(n)leq 10(k-1)+d$$
and
$$P(n)leq d9^{k-1},$$
so
$$d10^{k-1}leq n=S(n)+P(n)leq dleft(9^{k-1}+1right)+10(k-1).$$
$$dleft(10^{k-1}-9^{k-1}-1right)leq 10(k-1).$$
$$10^{k-1}-9^{k-1}-1leq 10(k-1).$$
Can you show that this has no solutions for $kgeq 4$, thus reducing it to the $k=3,d=1$ case? And then can you solve it from there using the same techniques you used in the body of your question?
answered Nov 24 at 5:22
Carl Schildkraut
11.1k11441
answered Nov 24 at 5:22
Carl Schildkraut
11.1k11441
answered Nov 24 at 5:22
Carl Schildkraut
11.1k11441
answered Nov 24 at 5:22
Carl Schildkraut
11.1k11441
11.1k11441
add a comment |
add a comment |
Hint:
Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$
Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...
answered Nov 24 at 9:48
greedoid
37.7k114794
add a comment |
Hint:
Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$
Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...
answered Nov 24 at 9:48
greedoid
37.7k114794
add a comment |
Hint:
Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$
Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...
answered Nov 24 at 9:48
greedoid
37.7k114794
Hint:
Write $$n= a_0+10a_1+10^2a_2+...+10^ka_k$$ where $a_kne 0$. Then $ngeq 10^k$, but $P(n)=a_0a_1...a_kleq 9^{k+1}$ and $S(n)leq 9(k+1)$. So we have $$10^kleq 9^{k+1}+9(k+1)$$
Clearly $k$ can not be big in fact only several $k$ works here (with calculator I find $kleq 20$)...
answered Nov 24 at 9:48
greedoid
37.7k114794
edited Nov 24 at 10:47
answered Nov 24 at 9:48
greedoid
37.7k114794
answered Nov 24 at 9:48
greedoid
37.7k114794
answered Nov 24 at 9:48
greedoid
37.7k114794
37.7k114794
add a comment |
add a comment |
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