Mixing
How do we find all possible values of $a$?
$begingroup$
$$2a5b equiv 0 mod 15$$
How do we find all possible values of $a$?
Here I tried to divide both sides by 2 and 5 respectively
$$ab equiv 0 mod 15 implies a in {3,6}, b in {0}$$
However, I think the way I used is so meaningless since this is equivalence, not an equation. Thus, I could not be able to take it from there.
Regards
modular-arithmetic
asked Jan 20 at 20:57
EnzoEnzo
19917
$endgroup$
add a comment |
$begingroup$
$$2a5b equiv 0 mod 15$$
How do we find all possible values of $a$?
Here I tried to divide both sides by 2 and 5 respectively
$$ab equiv 0 mod 15 implies a in {3,6}, b in {0}$$
However, I think the way I used is so meaningless since this is equivalence, not an equation. Thus, I could not be able to take it from there.
Regards
modular-arithmetic
asked Jan 20 at 20:57
EnzoEnzo
19917
$endgroup$
1
$begingroup$
If $$10abequiv 0 pmod {15}$$, then $$2abequiv 0 pmod 3$$ right?
$endgroup$
– Rhys Hughes
Jan 20 at 21:00
$begingroup$
How did you do that?
$endgroup$
– Enzo
Jan 20 at 21:00
$begingroup$
divided both sides by $5$
$endgroup$
– Rhys Hughes
Jan 20 at 21:06
$begingroup$
By $2a5b$, do you mean the number with digits $2,a,5,b$ or the product $2cdot a cdot 5 cdot b$?
$endgroup$
– bruderjakob17
Jan 20 at 21:17
add a comment |
$begingroup$
$$2a5b equiv 0 mod 15$$
How do we find all possible values of $a$?
Here I tried to divide both sides by 2 and 5 respectively
$$ab equiv 0 mod 15 implies a in {3,6}, b in {0}$$
However, I think the way I used is so meaningless since this is equivalence, not an equation. Thus, I could not be able to take it from there.
Regards
modular-arithmetic
asked Jan 20 at 20:57
EnzoEnzo
19917
$endgroup$
$$2a5b equiv 0 mod 15$$
How do we find all possible values of $a$?
Here I tried to divide both sides by 2 and 5 respectively
$$ab equiv 0 mod 15 implies a in {3,6}, b in {0}$$
However, I think the way I used is so meaningless since this is equivalence, not an equation. Thus, I could not be able to take it from there.
Regards
modular-arithmetic
modular-arithmetic
asked Jan 20 at 20:57
EnzoEnzo
19917
asked Jan 20 at 20:57
EnzoEnzo
19917
asked Jan 20 at 20:57
EnzoEnzo
19917
asked Jan 20 at 20:57
EnzoEnzo
19917
asked Jan 20 at 20:57
EnzoEnzo
19917
19917
1
$begingroup$
If $$10abequiv 0 pmod {15}$$, then $$2abequiv 0 pmod 3$$ right?
$endgroup$
– Rhys Hughes
Jan 20 at 21:00
$begingroup$
How did you do that?
$endgroup$
– Enzo
Jan 20 at 21:00
$begingroup$
divided both sides by $5$
$endgroup$
– Rhys Hughes
Jan 20 at 21:06
$begingroup$
By $2a5b$, do you mean the number with digits $2,a,5,b$ or the product $2cdot a cdot 5 cdot b$?
$endgroup$
– bruderjakob17
Jan 20 at 21:17
add a comment |
1
$begingroup$
If $$10abequiv 0 pmod {15}$$, then $$2abequiv 0 pmod 3$$ right?
$endgroup$
– Rhys Hughes
Jan 20 at 21:00
$begingroup$
How did you do that?
$endgroup$
– Enzo
Jan 20 at 21:00
$begingroup$
divided both sides by $5$
$endgroup$
– Rhys Hughes
Jan 20 at 21:06
$begingroup$
By $2a5b$, do you mean the number with digits $2,a,5,b$ or the product $2cdot a cdot 5 cdot b$?
$endgroup$
– bruderjakob17
Jan 20 at 21:17
1
1
$begingroup$
If $$10abequiv 0 pmod {15}$$, then $$2abequiv 0 pmod 3$$ right?
$endgroup$
– Rhys Hughes
Jan 20 at 21:00
$begingroup$
If $$10abequiv 0 pmod {15}$$, then $$2abequiv 0 pmod 3$$ right?
$endgroup$
– Rhys Hughes
Jan 20 at 21:00
$begingroup$
How did you do that?
$endgroup$
– Enzo
Jan 20 at 21:00
$begingroup$
How did you do that?
$endgroup$
– Enzo
Jan 20 at 21:00
$begingroup$
divided both sides by $5$
$endgroup$
– Rhys Hughes
Jan 20 at 21:06
$begingroup$
divided both sides by $5$
$endgroup$
– Rhys Hughes
Jan 20 at 21:06
$begingroup$
By $2a5b$, do you mean the number with digits $2,a,5,b$ or the product $2cdot a cdot 5 cdot b$?
$endgroup$
– bruderjakob17
Jan 20 at 21:17
$begingroup$
By $2a5b$, do you mean the number with digits $2,a,5,b$ or the product $2cdot a cdot 5 cdot b$?
$endgroup$
– bruderjakob17
Jan 20 at 21:17
add a comment |
1 Answer
1
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$begingroup$
Hint:
If $a,b$ are digits of the number $2a5b$ as I suppose, this number must be:
divisible by $5quad rightarrow quad b=5$ or $b=0$
and
divisible by $3 quad rightarrow quad 2+a+5+b$ is a multiple of $3$
can you do from this?
answered Jan 20 at 21:14
Emilio NovatiEmilio Novati
52.2k43474
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint:
If $a,b$ are digits of the number $2a5b$ as I suppose, this number must be:
divisible by $5quad rightarrow quad b=5$ or $b=0$
and
divisible by $3 quad rightarrow quad 2+a+5+b$ is a multiple of $3$
can you do from this?
answered Jan 20 at 21:14
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
add a comment |
$begingroup$
Hint:
If $a,b$ are digits of the number $2a5b$ as I suppose, this number must be:
divisible by $5quad rightarrow quad b=5$ or $b=0$
and
divisible by $3 quad rightarrow quad 2+a+5+b$ is a multiple of $3$
can you do from this?
answered Jan 20 at 21:14
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
add a comment |
$begingroup$
Hint:
If $a,b$ are digits of the number $2a5b$ as I suppose, this number must be:
divisible by $5quad rightarrow quad b=5$ or $b=0$
and
divisible by $3 quad rightarrow quad 2+a+5+b$ is a multiple of $3$
can you do from this?
answered Jan 20 at 21:14
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
Hint:
If $a,b$ are digits of the number $2a5b$ as I suppose, this number must be:
divisible by $5quad rightarrow quad b=5$ or $b=0$
and
divisible by $3 quad rightarrow quad 2+a+5+b$ is a multiple of $3$
can you do from this?
answered Jan 20 at 21:14
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 20 at 21:14
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 20 at 21:14
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 20 at 21:14
Emilio NovatiEmilio Novati
52.2k43474
52.2k43474
add a comment |
add a comment |
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1
$begingroup$
If $$10abequiv 0 pmod {15}$$, then $$2abequiv 0 pmod 3$$ right?
$endgroup$
– Rhys Hughes
Jan 20 at 21:00
$begingroup$
How did you do that?
$endgroup$
– Enzo
Jan 20 at 21:00
$begingroup$
divided both sides by $5$
$endgroup$
– Rhys Hughes
Jan 20 at 21:06
$begingroup$
By $2a5b$, do you mean the number with digits $2,a,5,b$ or the product $2cdot a cdot 5 cdot b$?
$endgroup$
– bruderjakob17
Jan 20 at 21:17