Proving finite bases for a Harshad number?
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I'm having some trouble with harshad numbers. How do I prove that the number 136 is Harshad only for the bases 2,3,4,5 and 9 and every multiple of 136 thereafter?
number-theory elementary-number-theory proof-writing proof-explanation
edited Jan 2 at 3:33
spaceisdarkgreen
33.4k21753
asked Jan 2 at 0:48
nahm8 fkn8nahm8 fkn8
184
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add a comment |
$begingroup$
I'm having some trouble with harshad numbers. How do I prove that the number 136 is Harshad only for the bases 2,3,4,5 and 9 and every multiple of 136 thereafter?
number-theory elementary-number-theory proof-writing proof-explanation
edited Jan 2 at 3:33
spaceisdarkgreen
33.4k21753
asked Jan 2 at 0:48
nahm8 fkn8nahm8 fkn8
184
$endgroup$
add a comment |
$begingroup$
I'm having some trouble with harshad numbers. How do I prove that the number 136 is Harshad only for the bases 2,3,4,5 and 9 and every multiple of 136 thereafter?
number-theory elementary-number-theory proof-writing proof-explanation
edited Jan 2 at 3:33
spaceisdarkgreen
33.4k21753
asked Jan 2 at 0:48
nahm8 fkn8nahm8 fkn8
184
$endgroup$
I'm having some trouble with harshad numbers. How do I prove that the number 136 is Harshad only for the bases 2,3,4,5 and 9 and every multiple of 136 thereafter?
number-theory elementary-number-theory proof-writing proof-explanation
number-theory elementary-number-theory proof-writing proof-explanation
edited Jan 2 at 3:33
spaceisdarkgreen
33.4k21753
asked Jan 2 at 0:48
nahm8 fkn8nahm8 fkn8
184
edited Jan 2 at 3:33
spaceisdarkgreen
33.4k21753
asked Jan 2 at 0:48
nahm8 fkn8nahm8 fkn8
184
edited Jan 2 at 3:33
spaceisdarkgreen
33.4k21753
edited Jan 2 at 3:33
spaceisdarkgreen
33.4k21753
edited Jan 2 at 3:33
spaceisdarkgreen
33.4k21753
33.4k21753
asked Jan 2 at 0:48
nahm8 fkn8nahm8 fkn8
184
asked Jan 2 at 0:48
nahm8 fkn8nahm8 fkn8
184
asked Jan 2 at 0:48
nahm8 fkn8nahm8 fkn8
184
184
add a comment |
add a comment |
1 Answer
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$begingroup$
Long solution:
If $n geq 12$ is any basis, then
$$136=a_0+a_1n$$
and
$$a_0+a_1| 136$$
Now for each $d |136$, meaning $d in {1, 2, 4, 8, 17, 34, 68, 136 }$ you can simply solve the system of equations
$$a_0+a_1n=136 \
a_0+a_1=d$$
by observing that
$$a_1(n-1)=136-d$$
leads to finitely many factorisations. You can eliminate many of them by observing that
$$136=a_0+a_1n geq a_1 cdot 12 Rightarrow
a_1 leq 11$$
The cases $n leq 11$ can be studied by simply writing the number $136$ out in each of these basis.
answered Jan 2 at 2:08
N. S.N. S.
104k7114209
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Long solution:
If $n geq 12$ is any basis, then
$$136=a_0+a_1n$$
and
$$a_0+a_1| 136$$
Now for each $d |136$, meaning $d in {1, 2, 4, 8, 17, 34, 68, 136 }$ you can simply solve the system of equations
$$a_0+a_1n=136 \
a_0+a_1=d$$
by observing that
$$a_1(n-1)=136-d$$
leads to finitely many factorisations. You can eliminate many of them by observing that
$$136=a_0+a_1n geq a_1 cdot 12 Rightarrow
a_1 leq 11$$
The cases $n leq 11$ can be studied by simply writing the number $136$ out in each of these basis.
answered Jan 2 at 2:08
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Long solution:
If $n geq 12$ is any basis, then
$$136=a_0+a_1n$$
and
$$a_0+a_1| 136$$
Now for each $d |136$, meaning $d in {1, 2, 4, 8, 17, 34, 68, 136 }$ you can simply solve the system of equations
$$a_0+a_1n=136 \
a_0+a_1=d$$
by observing that
$$a_1(n-1)=136-d$$
leads to finitely many factorisations. You can eliminate many of them by observing that
$$136=a_0+a_1n geq a_1 cdot 12 Rightarrow
a_1 leq 11$$
The cases $n leq 11$ can be studied by simply writing the number $136$ out in each of these basis.
answered Jan 2 at 2:08
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Long solution:
If $n geq 12$ is any basis, then
$$136=a_0+a_1n$$
and
$$a_0+a_1| 136$$
Now for each $d |136$, meaning $d in {1, 2, 4, 8, 17, 34, 68, 136 }$ you can simply solve the system of equations
$$a_0+a_1n=136 \
a_0+a_1=d$$
by observing that
$$a_1(n-1)=136-d$$
leads to finitely many factorisations. You can eliminate many of them by observing that
$$136=a_0+a_1n geq a_1 cdot 12 Rightarrow
a_1 leq 11$$
The cases $n leq 11$ can be studied by simply writing the number $136$ out in each of these basis.
answered Jan 2 at 2:08
N. S.N. S.
104k7114209
$endgroup$
Long solution:
If $n geq 12$ is any basis, then
$$136=a_0+a_1n$$
and
$$a_0+a_1| 136$$
Now for each $d |136$, meaning $d in {1, 2, 4, 8, 17, 34, 68, 136 }$ you can simply solve the system of equations
$$a_0+a_1n=136 \
a_0+a_1=d$$
by observing that
$$a_1(n-1)=136-d$$
leads to finitely many factorisations. You can eliminate many of them by observing that
$$136=a_0+a_1n geq a_1 cdot 12 Rightarrow
a_1 leq 11$$
The cases $n leq 11$ can be studied by simply writing the number $136$ out in each of these basis.
answered Jan 2 at 2:08
N. S.N. S.
104k7114209
answered Jan 2 at 2:08
N. S.N. S.
104k7114209
answered Jan 2 at 2:08
N. S.N. S.
104k7114209
answered Jan 2 at 2:08
N. S.N. S.
104k7114209
104k7114209
add a comment |
add a comment |
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