The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
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The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
On the condition that a number can be divided by $6$, it must be divisible by both $2$ and $3$
$$AABB equiv 0 pmod2tag1$$
$$AABB equiv 0 pmod3tag2$$
Reducing $AABB equiv 0 pmod2$,
$$AABB equiv 0 pmod2 implies 1000A + 100A + 10B + B implies B equiv 0 pmod2$$
Now reducing $AABB equiv 0 pmod3$,
$$AABB equiv 0 pmod3 implies 1000A + 100A +10B + B implies A + A + B + B equiv 0 pmod3 implies 2(A+B) equiv 0 pmod3$$
I think I've gone wrong so far.
modular-arithmetic divisibility
asked Sep 2 at 7:32
Busi
31818
 |Â
show 4 more comments
up vote
1
down vote
favorite
The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
On the condition that a number can be divided by $6$, it must be divisible by both $2$ and $3$
$$AABB equiv 0 pmod2tag1$$
$$AABB equiv 0 pmod3tag2$$
Reducing $AABB equiv 0 pmod2$,
$$AABB equiv 0 pmod2 implies 1000A + 100A + 10B + B implies B equiv 0 pmod2$$
Now reducing $AABB equiv 0 pmod3$,
$$AABB equiv 0 pmod3 implies 1000A + 100A +10B + B implies A + A + B + B equiv 0 pmod3 implies 2(A+B) equiv 0 pmod3$$
I think I've gone wrong so far.
modular-arithmetic divisibility
asked Sep 2 at 7:32
Busi
31818
1
Should it be $B equiv 0 pmod 2$?
– xbh
Sep 2 at 7:36
@xbh Yes, it should. Sorry for that mistake. See the new edit.
– Busi
Sep 2 at 7:37
$AABB=11times A0B$.
– Lord Shark the Unknown
Sep 2 at 7:39
1
It should be $2(A+B)equiv0pmod3$ You have $Bequiv0pmod2$ That does not say that $equiv0pmod3$
– saulspatz
Sep 2 at 7:40
@saulspatz How?
– Busi
Sep 2 at 7:44
 |Â
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
On the condition that a number can be divided by $6$, it must be divisible by both $2$ and $3$
$$AABB equiv 0 pmod2tag1$$
$$AABB equiv 0 pmod3tag2$$
Reducing $AABB equiv 0 pmod2$,
$$AABB equiv 0 pmod2 implies 1000A + 100A + 10B + B implies B equiv 0 pmod2$$
Now reducing $AABB equiv 0 pmod3$,
$$AABB equiv 0 pmod3 implies 1000A + 100A +10B + B implies A + A + B + B equiv 0 pmod3 implies 2(A+B) equiv 0 pmod3$$
I think I've gone wrong so far.
modular-arithmetic divisibility
asked Sep 2 at 7:32
Busi
31818
The four digit number $AABB$ can be divided by $6$ without remainder. Determine the greatest and least value of $AABB$.
On the condition that a number can be divided by $6$, it must be divisible by both $2$ and $3$
$$AABB equiv 0 pmod2tag1$$
$$AABB equiv 0 pmod3tag2$$
Reducing $AABB equiv 0 pmod2$,
$$AABB equiv 0 pmod2 implies 1000A + 100A + 10B + B implies B equiv 0 pmod2$$
Now reducing $AABB equiv 0 pmod3$,
$$AABB equiv 0 pmod3 implies 1000A + 100A +10B + B implies A + A + B + B equiv 0 pmod3 implies 2(A+B) equiv 0 pmod3$$
I think I've gone wrong so far.
modular-arithmetic divisibility
modular-arithmetic divisibility
asked Sep 2 at 7:32
Busi
31818
asked Sep 2 at 7:32
Busi
31818
edited Sep 2 at 7:45
asked Sep 2 at 7:32
Busi
31818
asked Sep 2 at 7:32
Busi
31818
asked Sep 2 at 7:32
Busi
31818
31818
1
Should it be $B equiv 0 pmod 2$?
– xbh
Sep 2 at 7:36
@xbh Yes, it should. Sorry for that mistake. See the new edit.
– Busi
Sep 2 at 7:37
$AABB=11times A0B$.
– Lord Shark the Unknown
Sep 2 at 7:39
1
It should be $2(A+B)equiv0pmod3$ You have $Bequiv0pmod2$ That does not say that $equiv0pmod3$
– saulspatz
Sep 2 at 7:40
@saulspatz How?
– Busi
Sep 2 at 7:44
 |Â
show 4 more comments
1
Should it be $B equiv 0 pmod 2$?
– xbh
Sep 2 at 7:36
@xbh Yes, it should. Sorry for that mistake. See the new edit.
– Busi
Sep 2 at 7:37
$AABB=11times A0B$.
– Lord Shark the Unknown
Sep 2 at 7:39
1
It should be $2(A+B)equiv0pmod3$ You have $Bequiv0pmod2$ That does not say that $equiv0pmod3$
– saulspatz
Sep 2 at 7:40
@saulspatz How?
– Busi
Sep 2 at 7:44
1
1
Should it be $B equiv 0 pmod 2$?
– xbh
Sep 2 at 7:36
Should it be $B equiv 0 pmod 2$?
– xbh
Sep 2 at 7:36
@xbh Yes, it should. Sorry for that mistake. See the new edit.
– Busi
Sep 2 at 7:37
@xbh Yes, it should. Sorry for that mistake. See the new edit.
– Busi
Sep 2 at 7:37
$AABB=11times A0B$.
– Lord Shark the Unknown
Sep 2 at 7:39
$AABB=11times A0B$.
– Lord Shark the Unknown
Sep 2 at 7:39
1
1
It should be $2(A+B)equiv0pmod3$ You have $Bequiv0pmod2$ That does not say that $equiv0pmod3$
– saulspatz
Sep 2 at 7:40
It should be $2(A+B)equiv0pmod3$ You have $Bequiv0pmod2$ That does not say that $equiv0pmod3$
– saulspatz
Sep 2 at 7:40
@saulspatz How?
– Busi
Sep 2 at 7:44
@saulspatz How?
– Busi
Sep 2 at 7:44
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You have shown that we must have $B$ even and $A+B$ divisible by $3$. To make $AABB$ as large as possible, we'd like to have $A=9$ If this is so, can we find an even number $B$ such that $3|(9+B)?$ Similarly, to make $AABB$ as small as possible, we'd like to have $A=1$. What are the choices for $B?$
answered Sep 2 at 7:59
saulspatz
11.6k21324
Isnt it $3|2(9+B)$, which implies that $3|18+2B$?
– Busi
Sep 2 at 8:04
To the proposer: If $nin Bbb Z$ then $3|2n iff 3|n.$ And $3|(9+B)iff 3|B.$ So $3|2(9+b)iff 3|(9+B)iff 3|B.$ So if $A=9$ then $Bin 0,3,6,9.$ But in this Q, $B$ must also be even. So if $A=9$ then $Bin 0,3,6,9cap 0,2,4,6,8=0,6.$
– DanielWainfleet
Sep 2 at 10:12
@Busi If $3|2n$ then $3|n$. If $p$ is prime and $p|ab$ then $p|a$ or $p|b$. Since $3nmid2$ it must be that $3|n$.
– saulspatz
Sep 2 at 14:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have shown that we must have $B$ even and $A+B$ divisible by $3$. To make $AABB$ as large as possible, we'd like to have $A=9$ If this is so, can we find an even number $B$ such that $3|(9+B)?$ Similarly, to make $AABB$ as small as possible, we'd like to have $A=1$. What are the choices for $B?$
answered Sep 2 at 7:59
saulspatz
11.6k21324
Isnt it $3|2(9+B)$, which implies that $3|18+2B$?
– Busi
Sep 2 at 8:04
To the proposer: If $nin Bbb Z$ then $3|2n iff 3|n.$ And $3|(9+B)iff 3|B.$ So $3|2(9+b)iff 3|(9+B)iff 3|B.$ So if $A=9$ then $Bin 0,3,6,9.$ But in this Q, $B$ must also be even. So if $A=9$ then $Bin 0,3,6,9cap 0,2,4,6,8=0,6.$
– DanielWainfleet
Sep 2 at 10:12
@Busi If $3|2n$ then $3|n$. If $p$ is prime and $p|ab$ then $p|a$ or $p|b$. Since $3nmid2$ it must be that $3|n$.
– saulspatz
Sep 2 at 14:21
add a comment |Â
up vote
1
down vote
accepted
You have shown that we must have $B$ even and $A+B$ divisible by $3$. To make $AABB$ as large as possible, we'd like to have $A=9$ If this is so, can we find an even number $B$ such that $3|(9+B)?$ Similarly, to make $AABB$ as small as possible, we'd like to have $A=1$. What are the choices for $B?$
answered Sep 2 at 7:59
saulspatz
11.6k21324
Isnt it $3|2(9+B)$, which implies that $3|18+2B$?
– Busi
Sep 2 at 8:04
To the proposer: If $nin Bbb Z$ then $3|2n iff 3|n.$ And $3|(9+B)iff 3|B.$ So $3|2(9+b)iff 3|(9+B)iff 3|B.$ So if $A=9$ then $Bin 0,3,6,9.$ But in this Q, $B$ must also be even. So if $A=9$ then $Bin 0,3,6,9cap 0,2,4,6,8=0,6.$
– DanielWainfleet
Sep 2 at 10:12
@Busi If $3|2n$ then $3|n$. If $p$ is prime and $p|ab$ then $p|a$ or $p|b$. Since $3nmid2$ it must be that $3|n$.
– saulspatz
Sep 2 at 14:21
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have shown that we must have $B$ even and $A+B$ divisible by $3$. To make $AABB$ as large as possible, we'd like to have $A=9$ If this is so, can we find an even number $B$ such that $3|(9+B)?$ Similarly, to make $AABB$ as small as possible, we'd like to have $A=1$. What are the choices for $B?$
answered Sep 2 at 7:59
saulspatz
11.6k21324
You have shown that we must have $B$ even and $A+B$ divisible by $3$. To make $AABB$ as large as possible, we'd like to have $A=9$ If this is so, can we find an even number $B$ such that $3|(9+B)?$ Similarly, to make $AABB$ as small as possible, we'd like to have $A=1$. What are the choices for $B?$
answered Sep 2 at 7:59
saulspatz
11.6k21324
answered Sep 2 at 7:59
saulspatz
11.6k21324
answered Sep 2 at 7:59
saulspatz
11.6k21324
answered Sep 2 at 7:59
saulspatz
11.6k21324
11.6k21324
Isnt it $3|2(9+B)$, which implies that $3|18+2B$?
– Busi
Sep 2 at 8:04
To the proposer: If $nin Bbb Z$ then $3|2n iff 3|n.$ And $3|(9+B)iff 3|B.$ So $3|2(9+b)iff 3|(9+B)iff 3|B.$ So if $A=9$ then $Bin 0,3,6,9.$ But in this Q, $B$ must also be even. So if $A=9$ then $Bin 0,3,6,9cap 0,2,4,6,8=0,6.$
– DanielWainfleet
Sep 2 at 10:12
@Busi If $3|2n$ then $3|n$. If $p$ is prime and $p|ab$ then $p|a$ or $p|b$. Since $3nmid2$ it must be that $3|n$.
– saulspatz
Sep 2 at 14:21
add a comment |Â
Isnt it $3|2(9+B)$, which implies that $3|18+2B$?
– Busi
Sep 2 at 8:04
To the proposer: If $nin Bbb Z$ then $3|2n iff 3|n.$ And $3|(9+B)iff 3|B.$ So $3|2(9+b)iff 3|(9+B)iff 3|B.$ So if $A=9$ then $Bin 0,3,6,9.$ But in this Q, $B$ must also be even. So if $A=9$ then $Bin 0,3,6,9cap 0,2,4,6,8=0,6.$
– DanielWainfleet
Sep 2 at 10:12
@Busi If $3|2n$ then $3|n$. If $p$ is prime and $p|ab$ then $p|a$ or $p|b$. Since $3nmid2$ it must be that $3|n$.
– saulspatz
Sep 2 at 14:21
Isnt it $3|2(9+B)$, which implies that $3|18+2B$?
– Busi
Sep 2 at 8:04
Isnt it $3|2(9+B)$, which implies that $3|18+2B$?
– Busi
Sep 2 at 8:04
To the proposer: If $nin Bbb Z$ then $3|2n iff 3|n.$ And $3|(9+B)iff 3|B.$ So $3|2(9+b)iff 3|(9+B)iff 3|B.$ So if $A=9$ then $Bin 0,3,6,9.$ But in this Q, $B$ must also be even. So if $A=9$ then $Bin 0,3,6,9cap 0,2,4,6,8=0,6.$
– DanielWainfleet
Sep 2 at 10:12
To the proposer: If $nin Bbb Z$ then $3|2n iff 3|n.$ And $3|(9+B)iff 3|B.$ So $3|2(9+b)iff 3|(9+B)iff 3|B.$ So if $A=9$ then $Bin 0,3,6,9.$ But in this Q, $B$ must also be even. So if $A=9$ then $Bin 0,3,6,9cap 0,2,4,6,8=0,6.$
– DanielWainfleet
Sep 2 at 10:12
@Busi If $3|2n$ then $3|n$. If $p$ is prime and $p|ab$ then $p|a$ or $p|b$. Since $3nmid2$ it must be that $3|n$.
– saulspatz
Sep 2 at 14:21
@Busi If $3|2n$ then $3|n$. If $p$ is prime and $p|ab$ then $p|a$ or $p|b$. Since $3nmid2$ it must be that $3|n$.
– saulspatz
Sep 2 at 14:21
add a comment |Â
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1
Should it be $B equiv 0 pmod 2$?
– xbh
Sep 2 at 7:36
@xbh Yes, it should. Sorry for that mistake. See the new edit.
– Busi
Sep 2 at 7:37
$AABB=11times A0B$.
– Lord Shark the Unknown
Sep 2 at 7:39
1
It should be $2(A+B)equiv0pmod3$ You have $Bequiv0pmod2$ That does not say that $equiv0pmod3$
– saulspatz
Sep 2 at 7:40
@saulspatz How?
– Busi
Sep 2 at 7:44