Prove that if $d$ is a divisor of $a$ and $b$ then $d$ is a divisor of $2020$, I think the book is wrong

Question

$a = 113n + 10$

$b = 89n - 10$

$n$ is a natural number

Question: Prove that if $d$ is a divisor of $a$ and $b$ then $d$ is a divisor of $2020$

The book's solution:

$d / a$

$d / b$

then $d / a + b$ so $d / 202n$

$d / 10(202n)$

$d / 2020n$

so $d / 2020$

But here is my problem with this solution.

$d / 2020n$ shouldn't mean $d / 2020$ because we don't know if $n$ is co-prime with $d$ or not.

If $n$ was co-prime then yes $d / 2020$ but we don't know that so I think the solution is wrong.

What do you think? Is the book correct and am I wrong?

Do you have a link to the book solution? This looks more like a student's solution. — Dietrich Burde, Jan 23 '22 at 17:22
See here in the linked dupe for the standard methods to solve such problems. — Bill Dubuque, Jan 23 '22 at 17:39

score 3 · Accepted Answer · answered Jan 23 '22 at 17:21

3

Yes $d | 2020 n$ doesn't mean that $d | 2020$.
May be in the book they want to say : $$\left\{\begin{array}{lcl} d & | & a \\ d & | & b \end{array}\right.$$ then : $$\left\{\begin{array}{lcl} d & | & 113 n + 10 \\ d & | & 89 n - 10 \end{array}\right.$$ then : $$d | 89 (113 n + 10) - 113 (89 n - 10) = 2020$$

answered Jan 23 '22 at 17:21

Essaidi

1,733

Consider using \mid instead of | to avoid spacing issues – Sil Jan 23 '22 at 17:27
Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jan 23 '22 at 17:41

Prove that if $d$ is a divisor of $a$ and $b$ then $d$ is a divisor of $2020$, I think the book is wrong

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