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My thinking:

$37j = 12k \rightarrow \:j+k=\frac{49j}{12}$

$\frac{49j}{12}$ = $7\left(\frac{7j}{12}\right)$

Here's where I'm stuck. How do I show that $\left(\frac{7j}{12}\right) \in \mathbb{N}$

Thank you.

someman112
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  • $!\bmod 7!:,\ \color{#0a0}{\overbrace{37}^{\large \equiv,\ 2}}j,\equiv!\color{#c00}{\overbrace{12}^{\large\equiv\ -2}}k!\iff \color{#0a0}2j\equiv \color{#c00}{-2}k!!\overset{\div\ 2}\iff j\equiv -k\iff j+k\equiv 0.,$ Note $,2,$ is coprime to the modulus $,7,$ so it is invertible so cancellable. See the linked dupe for general methods for solving linear congruences (above we solved for $,j,). \ $ – Bill Dubuque Mar 30 '22 at 08:17
  • I haven't learned mod yet in my class. I'm supposed to do it without that. – someman112 Mar 31 '22 at 03:43
  • The idea of the above congruence proof is simple: $,2(j!+!k),$ and $,37j!-!12k,$ differ by a multiple of $7$, viz. $7(5j!-!2k),,$ so since $7$ divides $,37j!-!12k,(= 0)$ it must also divide $,2(j!+!k),,$ so $,7\mid j!+!k,,$ by $,7\mid 2n\Rightarrow 7\mid n,,$ by $,(7,2)=1,$ & Euclid's Lemma (or directly: $,7\mid 4(2n)!-!7n = n)$. Congruence language simplifies arguments about divisibility relations by transforming then into more intuitive equation arithmetic, e.g. in my prior comment we solved a linear equation – Bill Dubuque Mar 31 '22 at 06:44
  • Directly: $\ j!+!k = 4:!(\overbrace{37j!-!12k}^{\large\color{#c00} 0})!-!\color{#c00}7^2(3j!-!k)$ is divisible by $\color{#c00}7$ since both summands are. $\ \ $ – Bill Dubuque Mar 31 '22 at 06:57
  • To complete the proof your way (using fractions) note that

    $$ \dfrac{37}{12} = \dfrac{k}j,\overset{+\ 1}\Longrightarrow, \dfrac{49}{12} =\dfrac{j+k}j\qquad $$

    but $,(49,12)=1,$ thus $,49/12,$ is reduced, therefore $,49,\mid, j+k\ \ $

     – Bill Dubuque Mar 31 '22 at 07:12

3 Answers3

0

We are given $37j=12k$, so $12\mid 37j$.

Observe that $(12,37)=1$. Then $12\mid 37j$ implies that $12\mid j$, so $j=12n$, for some $n\in \mathbb{N}$. It is easy to see then that $\frac{7j}{12}\in \mathbb{N}$.

Hanul Jeon
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neophyte
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-1

If $37j=12k$, then $k=37k_1$ and $j=12j_1$ for some $k_1,j_1\in \mathbb{N}$. Then try to show that $k_1=j_1$ and hence $k+j=49k_1$ and you are done.

Moreover, we have shown that $49\mid k+j$.

RFZ
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-1

Given that $j$ and $k$ are natural numbers satisfying the relation $$ 37 j = 12 k $$

This shows that $j = 12 m$ and $k = 37 m$ for some $m \in \mathbf{N}$.

Now, we note that $$ j + k = 12 m + 37 m = 49 m \tag{1} $$

By division algorithm, it is immediate from (1) that $49$ is a divisor of $j + k$.

We have shown that $49 \ | \ j + k$.

Dr. Sundar
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