Mixing
Tough Divisibility Problem
Clash Royale CLAN TAG#URR8PPP
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When the five digit number $2A13B$ is divided by $19$, the remainder is $12$. Determine the remainder of $3A21B$ when divided by $19$.
$$2A13B equiv 12 pmod19$$
$$20000 + 1000A + 100 + 30 + B equiv 12 pmod19$$
$$ 5 + 12A + 5 + 11 + B equiv 12 pmod19$$
$$ 21+ 12A+ B equiv 12 pmod19$$
$$ 12A+ B + 9 equiv 0 pmod19$$
This is where I'm stuck.
modular-arithmetic divisibility
edited Sep 1 at 17:10
José Carlos Santos
120k16101182
asked Sep 1 at 15:31
Busi
31118
add a comment |Â
up vote
3
down vote
favorite
When the five digit number $2A13B$ is divided by $19$, the remainder is $12$. Determine the remainder of $3A21B$ when divided by $19$.
$$2A13B equiv 12 pmod19$$
$$20000 + 1000A + 100 + 30 + B equiv 12 pmod19$$
$$ 5 + 12A + 5 + 11 + B equiv 12 pmod19$$
$$ 21+ 12A+ B equiv 12 pmod19$$
$$ 12A+ B + 9 equiv 0 pmod19$$
This is where I'm stuck.
modular-arithmetic divisibility
edited Sep 1 at 17:10
José Carlos Santos
120k16101182
asked Sep 1 at 15:31
Busi
31118
Can you reduce $3000 + 1000A + 210 + B$ modulo 19 in the same way?
– Matthew Leingang
Sep 1 at 15:34
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
When the five digit number $2A13B$ is divided by $19$, the remainder is $12$. Determine the remainder of $3A21B$ when divided by $19$.
$$2A13B equiv 12 pmod19$$
$$20000 + 1000A + 100 + 30 + B equiv 12 pmod19$$
$$ 5 + 12A + 5 + 11 + B equiv 12 pmod19$$
$$ 21+ 12A+ B equiv 12 pmod19$$
$$ 12A+ B + 9 equiv 0 pmod19$$
This is where I'm stuck.
modular-arithmetic divisibility
edited Sep 1 at 17:10
José Carlos Santos
120k16101182
asked Sep 1 at 15:31
Busi
31118
When the five digit number $2A13B$ is divided by $19$, the remainder is $12$. Determine the remainder of $3A21B$ when divided by $19$.
$$2A13B equiv 12 pmod19$$
$$20000 + 1000A + 100 + 30 + B equiv 12 pmod19$$
$$ 5 + 12A + 5 + 11 + B equiv 12 pmod19$$
$$ 21+ 12A+ B equiv 12 pmod19$$
$$ 12A+ B + 9 equiv 0 pmod19$$
This is where I'm stuck.
modular-arithmetic divisibility
edited Sep 1 at 17:10
José Carlos Santos
120k16101182
asked Sep 1 at 15:31
Busi
31118
edited Sep 1 at 17:10
José Carlos Santos
120k16101182
edited Sep 1 at 17:10
José Carlos Santos
120k16101182
edited Sep 1 at 17:10
José Carlos Santos
120k16101182
120k16101182
asked Sep 1 at 15:31
Busi
31118
asked Sep 1 at 15:31
Busi
31118
asked Sep 1 at 15:31
Busi
31118
31118
Can you reduce $3000 + 1000A + 210 + B$ modulo 19 in the same way?
– Matthew Leingang
Sep 1 at 15:34
add a comment |Â
Can you reduce $3000 + 1000A + 210 + B$ modulo 19 in the same way?
– Matthew Leingang
Sep 1 at 15:34
Can you reduce $3000 + 1000A + 210 + B$ modulo 19 in the same way?
– Matthew Leingang
Sep 1 at 15:34
Can you reduce $3000 + 1000A + 210 + B$ modulo 19 in the same way?
– Matthew Leingang
Sep 1 at 15:34
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
You have some errors, which I will fix:
$$beginalignoverline2A13B=20000 + 1000A + 100 + 30 + B &equiv 12 pmod19 Rightarrow \
(19cdot 1052+12)+(19cdot 52+12)A+(19cdot 5+5)+(19cdot 1+11)+B&equiv 12 pmod19 Rightarrow \
12+12A+5+11+B&equiv 12 pmod19 Rightarrow \
12A+19cdot 1+9+B&equiv 12 pmod19 Rightarrow \
12A+9+B&equiv 12 pmod19 Rightarrow \
12A+B&equiv 3pmod19.endalign$$
Since $0le A,Ble 9$, then: $(A,B)=(0,3),(3,5),(6,7),(9,9)$.
Similarly:
$$beginalignoverline3A21B=30000 + 1000A + 200 + 10 + B &equiv x pmod19 Rightarrow \
(19cdot 1578+18)+(19cdot 52+12)A+(19cdot 10+10)+10+B&equiv x pmod19 Rightarrow \
18+12A+10+10+B&equiv x pmod19 Rightarrow \
12A+19cdot 2+B&equiv x pmod19 Rightarrow \
12A+B&equiv x pmod19.endalign$$
So, $x=3$.
For example, take $overline2A13B=20133equiv 12 pmod19$ and $overline3A21B=30213equiv 3pmod19$.
answered Sep 1 at 17:25
farruhota
15k2734
add a comment |Â
up vote
11
down vote
Hint: $3A21B = 2A13B + 10000 + 80$
answered Sep 1 at 15:39
user96233
79545
3
Yes, everybody seems to have missed that!
– TonyK
Sep 1 at 18:40
3
This is by far the best answer, and arguably the only reasonable one to the problem. All the others are making it ridiculously overcomplicated.
– R..
Sep 2 at 1:25
add a comment |Â
up vote
2
down vote
Hint : $B equiv -12A-9 pmod19$. Next, in $3A21B pmod19$ you can replace $B$ by the RHS expression for it.
edited Sep 1 at 15:55
Clayton
18.3k22883
answered Sep 1 at 15:34
Ewan Delanoy
40.9k440102
add a comment |Â
up vote
2
down vote
beginalign30,000+1,000A+200+10+Bequiv xpmod19&iff-1+12A+10+10+Bequiv xpmod19\&iff12A+Bequiv xpmod19.endalignTherefore, since $12A+B+9equiv0pmod19$, take $x=10$.
answered Sep 1 at 15:41
José Carlos Santos
120k16101182
I didn't get your answer properly. On the condition that $12A+Bequiv9pmod19$, we get $12A + B equiv x pmod19 implies 12A+Bequiv9pmod19$. Does it denote the right answer is $9$. You had also written "take $x = 10$", why?
– Busi
Sep 1 at 17:11
@Busi There was a typo in my answer, which I have already edited. I wrote $12A+Bequiv9pmod19$ where I should have written $12A+B+9equiv0pmod19$. Thank you.
– José Carlos Santos
Sep 1 at 17:16
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have some errors, which I will fix:
$$beginalignoverline2A13B=20000 + 1000A + 100 + 30 + B &equiv 12 pmod19 Rightarrow \
(19cdot 1052+12)+(19cdot 52+12)A+(19cdot 5+5)+(19cdot 1+11)+B&equiv 12 pmod19 Rightarrow \
12+12A+5+11+B&equiv 12 pmod19 Rightarrow \
12A+19cdot 1+9+B&equiv 12 pmod19 Rightarrow \
12A+9+B&equiv 12 pmod19 Rightarrow \
12A+B&equiv 3pmod19.endalign$$
Since $0le A,Ble 9$, then: $(A,B)=(0,3),(3,5),(6,7),(9,9)$.
Similarly:
$$beginalignoverline3A21B=30000 + 1000A + 200 + 10 + B &equiv x pmod19 Rightarrow \
(19cdot 1578+18)+(19cdot 52+12)A+(19cdot 10+10)+10+B&equiv x pmod19 Rightarrow \
18+12A+10+10+B&equiv x pmod19 Rightarrow \
12A+19cdot 2+B&equiv x pmod19 Rightarrow \
12A+B&equiv x pmod19.endalign$$
So, $x=3$.
For example, take $overline2A13B=20133equiv 12 pmod19$ and $overline3A21B=30213equiv 3pmod19$.
answered Sep 1 at 17:25
farruhota
15k2734
add a comment |Â
up vote
2
down vote
accepted
You have some errors, which I will fix:
$$beginalignoverline2A13B=20000 + 1000A + 100 + 30 + B &equiv 12 pmod19 Rightarrow \
(19cdot 1052+12)+(19cdot 52+12)A+(19cdot 5+5)+(19cdot 1+11)+B&equiv 12 pmod19 Rightarrow \
12+12A+5+11+B&equiv 12 pmod19 Rightarrow \
12A+19cdot 1+9+B&equiv 12 pmod19 Rightarrow \
12A+9+B&equiv 12 pmod19 Rightarrow \
12A+B&equiv 3pmod19.endalign$$
Since $0le A,Ble 9$, then: $(A,B)=(0,3),(3,5),(6,7),(9,9)$.
Similarly:
$$beginalignoverline3A21B=30000 + 1000A + 200 + 10 + B &equiv x pmod19 Rightarrow \
(19cdot 1578+18)+(19cdot 52+12)A+(19cdot 10+10)+10+B&equiv x pmod19 Rightarrow \
18+12A+10+10+B&equiv x pmod19 Rightarrow \
12A+19cdot 2+B&equiv x pmod19 Rightarrow \
12A+B&equiv x pmod19.endalign$$
So, $x=3$.
For example, take $overline2A13B=20133equiv 12 pmod19$ and $overline3A21B=30213equiv 3pmod19$.
answered Sep 1 at 17:25
farruhota
15k2734
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have some errors, which I will fix:
$$beginalignoverline2A13B=20000 + 1000A + 100 + 30 + B &equiv 12 pmod19 Rightarrow \
(19cdot 1052+12)+(19cdot 52+12)A+(19cdot 5+5)+(19cdot 1+11)+B&equiv 12 pmod19 Rightarrow \
12+12A+5+11+B&equiv 12 pmod19 Rightarrow \
12A+19cdot 1+9+B&equiv 12 pmod19 Rightarrow \
12A+9+B&equiv 12 pmod19 Rightarrow \
12A+B&equiv 3pmod19.endalign$$
Since $0le A,Ble 9$, then: $(A,B)=(0,3),(3,5),(6,7),(9,9)$.
Similarly:
$$beginalignoverline3A21B=30000 + 1000A + 200 + 10 + B &equiv x pmod19 Rightarrow \
(19cdot 1578+18)+(19cdot 52+12)A+(19cdot 10+10)+10+B&equiv x pmod19 Rightarrow \
18+12A+10+10+B&equiv x pmod19 Rightarrow \
12A+19cdot 2+B&equiv x pmod19 Rightarrow \
12A+B&equiv x pmod19.endalign$$
So, $x=3$.
For example, take $overline2A13B=20133equiv 12 pmod19$ and $overline3A21B=30213equiv 3pmod19$.
answered Sep 1 at 17:25
farruhota
15k2734
You have some errors, which I will fix:
$$beginalignoverline2A13B=20000 + 1000A + 100 + 30 + B &equiv 12 pmod19 Rightarrow \
(19cdot 1052+12)+(19cdot 52+12)A+(19cdot 5+5)+(19cdot 1+11)+B&equiv 12 pmod19 Rightarrow \
12+12A+5+11+B&equiv 12 pmod19 Rightarrow \
12A+19cdot 1+9+B&equiv 12 pmod19 Rightarrow \
12A+9+B&equiv 12 pmod19 Rightarrow \
12A+B&equiv 3pmod19.endalign$$
Since $0le A,Ble 9$, then: $(A,B)=(0,3),(3,5),(6,7),(9,9)$.
Similarly:
$$beginalignoverline3A21B=30000 + 1000A + 200 + 10 + B &equiv x pmod19 Rightarrow \
(19cdot 1578+18)+(19cdot 52+12)A+(19cdot 10+10)+10+B&equiv x pmod19 Rightarrow \
18+12A+10+10+B&equiv x pmod19 Rightarrow \
12A+19cdot 2+B&equiv x pmod19 Rightarrow \
12A+B&equiv x pmod19.endalign$$
So, $x=3$.
For example, take $overline2A13B=20133equiv 12 pmod19$ and $overline3A21B=30213equiv 3pmod19$.
answered Sep 1 at 17:25
farruhota
15k2734
answered Sep 1 at 17:25
farruhota
15k2734
answered Sep 1 at 17:25
farruhota
15k2734
answered Sep 1 at 17:25
farruhota
15k2734
15k2734
add a comment |Â
add a comment |Â
up vote
11
down vote
Hint: $3A21B = 2A13B + 10000 + 80$
answered Sep 1 at 15:39
user96233
79545
3
Yes, everybody seems to have missed that!
– TonyK
Sep 1 at 18:40
3
This is by far the best answer, and arguably the only reasonable one to the problem. All the others are making it ridiculously overcomplicated.
– R..
Sep 2 at 1:25
add a comment |Â
up vote
11
down vote
Hint: $3A21B = 2A13B + 10000 + 80$
answered Sep 1 at 15:39
user96233
79545
3
Yes, everybody seems to have missed that!
– TonyK
Sep 1 at 18:40
3
This is by far the best answer, and arguably the only reasonable one to the problem. All the others are making it ridiculously overcomplicated.
– R..
Sep 2 at 1:25
add a comment |Â
up vote
11
down vote
up vote
11
down vote
Hint: $3A21B = 2A13B + 10000 + 80$
answered Sep 1 at 15:39
user96233
79545
Hint: $3A21B = 2A13B + 10000 + 80$
answered Sep 1 at 15:39
user96233
79545
answered Sep 1 at 15:39
user96233
79545
answered Sep 1 at 15:39
user96233
79545
answered Sep 1 at 15:39
user96233
79545
79545
3
Yes, everybody seems to have missed that!
– TonyK
Sep 1 at 18:40
3
This is by far the best answer, and arguably the only reasonable one to the problem. All the others are making it ridiculously overcomplicated.
– R..
Sep 2 at 1:25
add a comment |Â
3
Yes, everybody seems to have missed that!
– TonyK
Sep 1 at 18:40
3
This is by far the best answer, and arguably the only reasonable one to the problem. All the others are making it ridiculously overcomplicated.
– R..
Sep 2 at 1:25
3
3
Yes, everybody seems to have missed that!
– TonyK
Sep 1 at 18:40
Yes, everybody seems to have missed that!
– TonyK
Sep 1 at 18:40
3
3
This is by far the best answer, and arguably the only reasonable one to the problem. All the others are making it ridiculously overcomplicated.
– R..
Sep 2 at 1:25
This is by far the best answer, and arguably the only reasonable one to the problem. All the others are making it ridiculously overcomplicated.
– R..
Sep 2 at 1:25
add a comment |Â
up vote
2
down vote
Hint : $B equiv -12A-9 pmod19$. Next, in $3A21B pmod19$ you can replace $B$ by the RHS expression for it.
edited Sep 1 at 15:55
Clayton
18.3k22883
answered Sep 1 at 15:34
Ewan Delanoy
40.9k440102
add a comment |Â
up vote
2
down vote
Hint : $B equiv -12A-9 pmod19$. Next, in $3A21B pmod19$ you can replace $B$ by the RHS expression for it.
edited Sep 1 at 15:55
Clayton
18.3k22883
answered Sep 1 at 15:34
Ewan Delanoy
40.9k440102
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint : $B equiv -12A-9 pmod19$. Next, in $3A21B pmod19$ you can replace $B$ by the RHS expression for it.
edited Sep 1 at 15:55
Clayton
18.3k22883
answered Sep 1 at 15:34
Ewan Delanoy
40.9k440102
Hint : $B equiv -12A-9 pmod19$. Next, in $3A21B pmod19$ you can replace $B$ by the RHS expression for it.
edited Sep 1 at 15:55
Clayton
18.3k22883
answered Sep 1 at 15:34
Ewan Delanoy
40.9k440102
edited Sep 1 at 15:55
Clayton
18.3k22883
edited Sep 1 at 15:55
Clayton
18.3k22883
edited Sep 1 at 15:55
Clayton
18.3k22883
18.3k22883
answered Sep 1 at 15:34
Ewan Delanoy
40.9k440102
answered Sep 1 at 15:34
Ewan Delanoy
40.9k440102
answered Sep 1 at 15:34
Ewan Delanoy
40.9k440102
40.9k440102
add a comment |Â
add a comment |Â
up vote
2
down vote
beginalign30,000+1,000A+200+10+Bequiv xpmod19&iff-1+12A+10+10+Bequiv xpmod19\&iff12A+Bequiv xpmod19.endalignTherefore, since $12A+B+9equiv0pmod19$, take $x=10$.
answered Sep 1 at 15:41
José Carlos Santos
120k16101182
I didn't get your answer properly. On the condition that $12A+Bequiv9pmod19$, we get $12A + B equiv x pmod19 implies 12A+Bequiv9pmod19$. Does it denote the right answer is $9$. You had also written "take $x = 10$", why?
– Busi
Sep 1 at 17:11
@Busi There was a typo in my answer, which I have already edited. I wrote $12A+Bequiv9pmod19$ where I should have written $12A+B+9equiv0pmod19$. Thank you.
– José Carlos Santos
Sep 1 at 17:16
add a comment |Â
up vote
2
down vote
beginalign30,000+1,000A+200+10+Bequiv xpmod19&iff-1+12A+10+10+Bequiv xpmod19\&iff12A+Bequiv xpmod19.endalignTherefore, since $12A+B+9equiv0pmod19$, take $x=10$.
answered Sep 1 at 15:41
José Carlos Santos
120k16101182
I didn't get your answer properly. On the condition that $12A+Bequiv9pmod19$, we get $12A + B equiv x pmod19 implies 12A+Bequiv9pmod19$. Does it denote the right answer is $9$. You had also written "take $x = 10$", why?
– Busi
Sep 1 at 17:11
@Busi There was a typo in my answer, which I have already edited. I wrote $12A+Bequiv9pmod19$ where I should have written $12A+B+9equiv0pmod19$. Thank you.
– José Carlos Santos
Sep 1 at 17:16
add a comment |Â
up vote
2
down vote
up vote
2
down vote
beginalign30,000+1,000A+200+10+Bequiv xpmod19&iff-1+12A+10+10+Bequiv xpmod19\&iff12A+Bequiv xpmod19.endalignTherefore, since $12A+B+9equiv0pmod19$, take $x=10$.
answered Sep 1 at 15:41
José Carlos Santos
120k16101182
beginalign30,000+1,000A+200+10+Bequiv xpmod19&iff-1+12A+10+10+Bequiv xpmod19\&iff12A+Bequiv xpmod19.endalignTherefore, since $12A+B+9equiv0pmod19$, take $x=10$.
answered Sep 1 at 15:41
José Carlos Santos
120k16101182
edited Sep 1 at 17:15
answered Sep 1 at 15:41
José Carlos Santos
120k16101182
answered Sep 1 at 15:41
José Carlos Santos
120k16101182
answered Sep 1 at 15:41
José Carlos Santos
120k16101182
120k16101182
I didn't get your answer properly. On the condition that $12A+Bequiv9pmod19$, we get $12A + B equiv x pmod19 implies 12A+Bequiv9pmod19$. Does it denote the right answer is $9$. You had also written "take $x = 10$", why?
– Busi
Sep 1 at 17:11
@Busi There was a typo in my answer, which I have already edited. I wrote $12A+Bequiv9pmod19$ where I should have written $12A+B+9equiv0pmod19$. Thank you.
– José Carlos Santos
Sep 1 at 17:16
add a comment |Â
I didn't get your answer properly. On the condition that $12A+Bequiv9pmod19$, we get $12A + B equiv x pmod19 implies 12A+Bequiv9pmod19$. Does it denote the right answer is $9$. You had also written "take $x = 10$", why?
– Busi
Sep 1 at 17:11
@Busi There was a typo in my answer, which I have already edited. I wrote $12A+Bequiv9pmod19$ where I should have written $12A+B+9equiv0pmod19$. Thank you.
– José Carlos Santos
Sep 1 at 17:16
I didn't get your answer properly. On the condition that $12A+Bequiv9pmod19$, we get $12A + B equiv x pmod19 implies 12A+Bequiv9pmod19$. Does it denote the right answer is $9$. You had also written "take $x = 10$", why?
– Busi
Sep 1 at 17:11
I didn't get your answer properly. On the condition that $12A+Bequiv9pmod19$, we get $12A + B equiv x pmod19 implies 12A+Bequiv9pmod19$. Does it denote the right answer is $9$. You had also written "take $x = 10$", why?
– Busi
Sep 1 at 17:11
@Busi There was a typo in my answer, which I have already edited. I wrote $12A+Bequiv9pmod19$ where I should have written $12A+B+9equiv0pmod19$. Thank you.
– José Carlos Santos
Sep 1 at 17:16
@Busi There was a typo in my answer, which I have already edited. I wrote $12A+Bequiv9pmod19$ where I should have written $12A+B+9equiv0pmod19$. Thank you.
– José Carlos Santos
Sep 1 at 17:16
add a comment |Â
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Can you reduce $3000 + 1000A + 210 + B$ modulo 19 in the same way?
– Matthew Leingang
Sep 1 at 15:34