In the subtraction sum on the right $a, b$ and $c$ are digits and $a$ is less than $b$. What is the value of...
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Here $bar{ba}=a+10b$ and $bar{ab}=b+10a$
Therefore $(a+10b)−(b+10a)=6+10c$
or $9(b−a)=6+10c$
or $c=displaystylefrac{9(b−a)−6}{10}$
Since $c$ is a digit, if by trial and error method we take $b−a=4$, then $c=3$
Therefore the value of $c$ is 3.
I am not satisfied with the approach. Is there any different approach and whether my procedure is correct? Please help.
linear-algebra
edited Nov 26 at 17:40
Mostafa Ayaz
13.5k3836
asked Nov 26 at 4:48
user3575652
87112
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up vote
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Here $bar{ba}=a+10b$ and $bar{ab}=b+10a$
Therefore $(a+10b)−(b+10a)=6+10c$
or $9(b−a)=6+10c$
or $c=displaystylefrac{9(b−a)−6}{10}$
Since $c$ is a digit, if by trial and error method we take $b−a=4$, then $c=3$
Therefore the value of $c$ is 3.
I am not satisfied with the approach. Is there any different approach and whether my procedure is correct? Please help.
linear-algebra
edited Nov 26 at 17:40
Mostafa Ayaz
13.5k3836
asked Nov 26 at 4:48
user3575652
87112
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here $bar{ba}=a+10b$ and $bar{ab}=b+10a$
Therefore $(a+10b)−(b+10a)=6+10c$
or $9(b−a)=6+10c$
or $c=displaystylefrac{9(b−a)−6}{10}$
Since $c$ is a digit, if by trial and error method we take $b−a=4$, then $c=3$
Therefore the value of $c$ is 3.
I am not satisfied with the approach. Is there any different approach and whether my procedure is correct? Please help.
linear-algebra
edited Nov 26 at 17:40
Mostafa Ayaz
13.5k3836
asked Nov 26 at 4:48
user3575652
87112
Here $bar{ba}=a+10b$ and $bar{ab}=b+10a$
Therefore $(a+10b)−(b+10a)=6+10c$
or $9(b−a)=6+10c$
or $c=displaystylefrac{9(b−a)−6}{10}$
Since $c$ is a digit, if by trial and error method we take $b−a=4$, then $c=3$
Therefore the value of $c$ is 3.
I am not satisfied with the approach. Is there any different approach and whether my procedure is correct? Please help.
linear-algebra
linear-algebra
edited Nov 26 at 17:40
Mostafa Ayaz
13.5k3836
asked Nov 26 at 4:48
user3575652
87112
edited Nov 26 at 17:40
Mostafa Ayaz
13.5k3836
asked Nov 26 at 4:48
user3575652
87112
edited Nov 26 at 17:40
Mostafa Ayaz
13.5k3836
edited Nov 26 at 17:40
Mostafa Ayaz
13.5k3836
edited Nov 26 at 17:40
Mostafa Ayaz
13.5k3836
13.5k3836
asked Nov 26 at 4:48
user3575652
87112
asked Nov 26 at 4:48
user3575652
87112
asked Nov 26 at 4:48
user3575652
87112
87112
add a comment |
add a comment |
2 Answers
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You have $(a+10b)-(b+10a)=9(b-a)$, which is divisible by $9$, so $c6$ has to be divisible by $9$. The usual divisibility test for $9$ says the sum of digits of $c6$ must be divisible by $9$, so $c+6=9, c=3$
answered Nov 26 at 4:54
Ross Millikan
289k23195367
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We must have that $$a-bequiv 6mod {10}$$and $bge a$. The only possibilities are $$(a,b)=(0,4)\(a,b)=(1,5)\(a,b)=(2,6)\(a,b)=(3,7)\(a,b)=(4,8)\(a,b)=(5,9)$$
answered Nov 26 at 17:44
Mostafa Ayaz
13.5k3836
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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votes
up vote
1
down vote
You have $(a+10b)-(b+10a)=9(b-a)$, which is divisible by $9$, so $c6$ has to be divisible by $9$. The usual divisibility test for $9$ says the sum of digits of $c6$ must be divisible by $9$, so $c+6=9, c=3$
answered Nov 26 at 4:54
Ross Millikan
289k23195367
add a comment |
up vote
1
down vote
You have $(a+10b)-(b+10a)=9(b-a)$, which is divisible by $9$, so $c6$ has to be divisible by $9$. The usual divisibility test for $9$ says the sum of digits of $c6$ must be divisible by $9$, so $c+6=9, c=3$
answered Nov 26 at 4:54
Ross Millikan
289k23195367
add a comment |
up vote
1
down vote
up vote
1
down vote
You have $(a+10b)-(b+10a)=9(b-a)$, which is divisible by $9$, so $c6$ has to be divisible by $9$. The usual divisibility test for $9$ says the sum of digits of $c6$ must be divisible by $9$, so $c+6=9, c=3$
answered Nov 26 at 4:54
Ross Millikan
289k23195367
You have $(a+10b)-(b+10a)=9(b-a)$, which is divisible by $9$, so $c6$ has to be divisible by $9$. The usual divisibility test for $9$ says the sum of digits of $c6$ must be divisible by $9$, so $c+6=9, c=3$
answered Nov 26 at 4:54
Ross Millikan
289k23195367
answered Nov 26 at 4:54
Ross Millikan
289k23195367
answered Nov 26 at 4:54
Ross Millikan
289k23195367
answered Nov 26 at 4:54
Ross Millikan
289k23195367
289k23195367
add a comment |
add a comment |
up vote
0
down vote
We must have that $$a-bequiv 6mod {10}$$and $bge a$. The only possibilities are $$(a,b)=(0,4)\(a,b)=(1,5)\(a,b)=(2,6)\(a,b)=(3,7)\(a,b)=(4,8)\(a,b)=(5,9)$$
answered Nov 26 at 17:44
Mostafa Ayaz
13.5k3836
add a comment |
up vote
0
down vote
We must have that $$a-bequiv 6mod {10}$$and $bge a$. The only possibilities are $$(a,b)=(0,4)\(a,b)=(1,5)\(a,b)=(2,6)\(a,b)=(3,7)\(a,b)=(4,8)\(a,b)=(5,9)$$
answered Nov 26 at 17:44
Mostafa Ayaz
13.5k3836
add a comment |
up vote
0
down vote
up vote
0
down vote
We must have that $$a-bequiv 6mod {10}$$and $bge a$. The only possibilities are $$(a,b)=(0,4)\(a,b)=(1,5)\(a,b)=(2,6)\(a,b)=(3,7)\(a,b)=(4,8)\(a,b)=(5,9)$$
answered Nov 26 at 17:44
Mostafa Ayaz
13.5k3836
We must have that $$a-bequiv 6mod {10}$$and $bge a$. The only possibilities are $$(a,b)=(0,4)\(a,b)=(1,5)\(a,b)=(2,6)\(a,b)=(3,7)\(a,b)=(4,8)\(a,b)=(5,9)$$
answered Nov 26 at 17:44
Mostafa Ayaz
13.5k3836
answered Nov 26 at 17:44
Mostafa Ayaz
13.5k3836
answered Nov 26 at 17:44
Mostafa Ayaz
13.5k3836
answered Nov 26 at 17:44
Mostafa Ayaz
13.5k3836
13.5k3836
add a comment |
add a comment |
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