If $ a > c $ , show that $ abc - cba = 99\left( {a - c} \right) $
Answer
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Hint: In order to show that $ abc - cba = 99\left( {a - c} \right) $ , assume that the $ abc $ and $ cba $ both are the three digit numbers. So write both the numbers in their expanded forms by identifying the once’s place, ten’s place and hundreds place. Subtract both of the expanded forms and combine all the like terms to get your required result.
Complete step by step solution:
We are given that the $ a > c $ , and we have to show that $ abc - cba = 99\left( {a - c} \right) $ .
So in order to show that $ abc - cba = 99\left( {a - c} \right) $ .
Let’s assume that the $ abc $ and $ cba $ are the three digits number.
In the number $ abc $ , the one’s place number is c , the ten’s place number is b and the hundred place number is a.
SO, writing the number abc in the expanded form.
Expanded form of a number is the sum of numbers according to their place values. The different place values of any number are Ones, Tens, Hundred, Thousands, Ten Thousand and so on.
a------🡪100a
b------🡪10b
c------🡪c
Hence, $ abc = 100a + 10b + c $ ----(1)
Now ,similarly with the other number $ cba $ , the one’s place number is a , the ten’s place number is b and the hundred place number is c.
So the expanded form of $ cba $ will be
$ cba = 100c + 10b + a $ -----(2)
Subtracting both of the expanded form with each other we get
$
abc - cba = 100a + 10b + c - \left( {100c + 10b + a} \right) \\
abc - cba = 100a + 10b + c - 100c - 10b - a \;
$
Combining all the like terms on the right-hand side of the equation ,
$
abc - cba = 100a + 10b + c - 100c - 10b - a \\
abc - cba = 99a - 99c \\
abc - cba = 99\left( {a - c} \right) \;
$
Therefore, we have successfully proved that $ abc - cba = 99\left( {a - c} \right) $
Note: 1.The name of the place values are according to the Indian system or to the International system of numeration.
2.Make sure the expansion of all the assumed three digits numbers are done correctly.
3. $ a,b,c $ in our question are some whole numbers.
4. Expansion of 5 digit number 72536 is $ 70000 + 2000 + 500 + 30 + 6 $ .
$ $
Complete step by step solution:
We are given that the $ a > c $ , and we have to show that $ abc - cba = 99\left( {a - c} \right) $ .
So in order to show that $ abc - cba = 99\left( {a - c} \right) $ .
Let’s assume that the $ abc $ and $ cba $ are the three digits number.
In the number $ abc $ , the one’s place number is c , the ten’s place number is b and the hundred place number is a.
SO, writing the number abc in the expanded form.
Expanded form of a number is the sum of numbers according to their place values. The different place values of any number are Ones, Tens, Hundred, Thousands, Ten Thousand and so on.
a------🡪100a
b------🡪10b
c------🡪c
Hence, $ abc = 100a + 10b + c $ ----(1)
Now ,similarly with the other number $ cba $ , the one’s place number is a , the ten’s place number is b and the hundred place number is c.
So the expanded form of $ cba $ will be
$ cba = 100c + 10b + a $ -----(2)
Subtracting both of the expanded form with each other we get
$
abc - cba = 100a + 10b + c - \left( {100c + 10b + a} \right) \\
abc - cba = 100a + 10b + c - 100c - 10b - a \;
$
Combining all the like terms on the right-hand side of the equation ,
$
abc - cba = 100a + 10b + c - 100c - 10b - a \\
abc - cba = 99a - 99c \\
abc - cba = 99\left( {a - c} \right) \;
$
Therefore, we have successfully proved that $ abc - cba = 99\left( {a - c} \right) $
Note: 1.The name of the place values are according to the Indian system or to the International system of numeration.
2.Make sure the expansion of all the assumed three digits numbers are done correctly.
3. $ a,b,c $ in our question are some whole numbers.
4. Expansion of 5 digit number 72536 is $ 70000 + 2000 + 500 + 30 + 6 $ .
$ $
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