Find all positive integers n such that 20n+2 can divide 2003n+2002

Question

Could someone give me a hint to this problem please? Here is what I do: $2003n+2002=100(20n+2)+3n+1802$ then I'm stuck at $20n+2|3n+1802$ Please help

Can you use Euclid's algorithm on the two numbers? – Toby Mak Mar 31 '20 at 04:53 — Toby Mak, Mar 31 '20 at 04:53

score 1 · Accepted Answer · answered Mar 31 '20 at 05:02

1

According to Euclid's algorithm, $\gcd(a, b) = \gcd(b, a \% b)$ where percent is modulo

$$20n+2|2003n+2002 \implies \gcd(20n+2,2003n+2002)=20n+2$$

The GCD equals

$$\gcd(20n+2, (2003n+2002)-100(20n+2))$$

$$=\gcd(20n+2,3n+1802)$$

$$=\gcd(3n+1802, 20n+2-6(3n+1802)$$

$$=\gcd(3n+1802, 2n-10810)$$

$$=\gcd(2n-10810, n+12612)$$

$$=\gcd(n+12612, -36034)=20n+2$$

Therefore $20n+2$ is a factor of $36034=2\times43\times419$ (positive)

$$20n+2=1, 2, 43, 86, 419, 838, 18017, 36034$$

As you can see, none of them end with "2" except 2. However since $n$ is positive, $20n+2\neq2$

Therefore, there're no positive integers $n$ that satisfy $20n+2|2003n+2002$.

answered Mar 31 '20 at 05:02

Gareth Ma

Yeah solution say there is really no n that satisfied it but the solution is complicated so I need other solution hahah Thanks so much – user635988 Mar 31 '20 at 05:21
Haha sure :)) good that it helps – Gareth Ma Mar 31 '20 at 05:22
@user635988 Hint: $ $ Simpler: modulo an odd prime common divisor we have $$\dfrac{2}{20}\equiv -n\equiv\dfrac{2002}{2003},\Rightarrow, 0\equiv 20(2002)-2(2003)\equiv 43\cdot 419\qquad$$ – Bill Dubuque Mar 31 '20 at 05:47

score 1 · Answer 2 · answered Mar 31 '20 at 06:05

1

If integer $d$ divides both $3n+1802, 20n+2$

$d$ must divide $20(3n+1807)-3(20n+2)=36134$

So, $20n+2$ must divide $36134$

$\iff10n+1$ divide $18067$ whose factors are $1,7,29,89,203,623,2581,18067$ none of which if of the form $10n+1$ for integer $n>0$

answered Mar 31 '20 at 06:05

lab bhattacharjee

i.e. eliminate $n$ as in proof $(1)$ in the linked dupe. – Bill Dubuque Mar 31 '20 at 06:09

2 Answers2