
The digits of a three digit number are consecutive odd numbers. The number is 51 less than 30 times the sum of its digits. What is the number?
(A) 975
(B) 579
(C) 759
(D) 597
Answer
587.7k+ views
Hint: Start with assuming the number to be some variable \[xyz\]. According to the question, two cases are possible. Either $x$ is maximum or $z$ is maximum. Solve both the cases separately and frame equations using the condition given in the question and solve them to find the possible values of $x,y$ and $z$ and hence the number.
Complete step-by-step answer:
Let the required three digit number is \[xyz\].
Now according to the question, $x,y$ and $z$ are consecutive odd numbers. So based on either $x$ is maximum or $z$ is maximum, two cases are possible.
Case 1 - If $x$ is maximum then we have $y = x - 2$ and $z = x - 4$ because they are consecutive odd numbers.
Further it is given that the number is 51 less than 30 times the sum of its digits. We know that the number \[xyz\] is written as $100x + 10y + z$. So from this we have:
\[ \Rightarrow 100x + 10y + z = 30\left( {x + y + z} \right) - 51\]
Putting the values of $y$ and $z$ we’ll get:
\[ \Rightarrow 100x + 10\left( {x - 2} \right) + \left( {x - 4} \right) = 30\left( {x + x - 2 + x - 4} \right) - 51\]
Solving it further for $x$, we’ll get:
\[
\Rightarrow 100x + 10x - 20 + x - 4 = 30\left( {3x - 6} \right) - 51 \\
\Rightarrow 111x - 24 = 90x - 180 - 51 \\
\Rightarrow 21x = - 207 \\
\Rightarrow x = - \dfrac{{207}}{{21}} = - \dfrac{{69}}{7} \\
\]
But since $x$ is a digit, its value cannot be negative and fractional. Therefore this case is not possible.
Case 1 - If $z$ is maximum then we have $y = x + 2$ and $z = x + 4$ because they are consecutive odd numbers.
Further it is given that the number is 51 less than 30 times the sum of its digits. We know that the number \[xyz\] is written as $100x + 10y + z$. So from this we have:
\[ \Rightarrow 100x + 10y + z = 30\left( {x + y + z} \right) - 51\]
Putting the values of $y$ and $z$ we’ll get:
\[ \Rightarrow 100x + 10\left( {x + 2} \right) + \left( {x + 4} \right) = 30\left( {x + x + 2 + x + 4} \right) - 51\]
Solving it further for $x$, we’ll get:
\[
\Rightarrow 100x + 10x + 20 + x + 4 = 30\left( {3x + 6} \right) - 51 \\
\Rightarrow 111x + 24 = 90x + 180 - 51 \\
\Rightarrow 21x = 105 \\
\Rightarrow x = 5 \\
\]
Thus the required value of $x$ is 5 and $y = x + 2 = 7,z = x + 4 = 9$
Hence the number \[xyz\] is 579. (B) is the correct option.
Note: Any three digit number \[xyz\] can be written as $100x + 10y + z$. Similarly any two digit number $xy$ can be written as $10x + y$. In the same manner, any $n$ digit number can be written as the sum of digits with the digits multiplied by their respective place value.
Complete step-by-step answer:
Let the required three digit number is \[xyz\].
Now according to the question, $x,y$ and $z$ are consecutive odd numbers. So based on either $x$ is maximum or $z$ is maximum, two cases are possible.
Case 1 - If $x$ is maximum then we have $y = x - 2$ and $z = x - 4$ because they are consecutive odd numbers.
Further it is given that the number is 51 less than 30 times the sum of its digits. We know that the number \[xyz\] is written as $100x + 10y + z$. So from this we have:
\[ \Rightarrow 100x + 10y + z = 30\left( {x + y + z} \right) - 51\]
Putting the values of $y$ and $z$ we’ll get:
\[ \Rightarrow 100x + 10\left( {x - 2} \right) + \left( {x - 4} \right) = 30\left( {x + x - 2 + x - 4} \right) - 51\]
Solving it further for $x$, we’ll get:
\[
\Rightarrow 100x + 10x - 20 + x - 4 = 30\left( {3x - 6} \right) - 51 \\
\Rightarrow 111x - 24 = 90x - 180 - 51 \\
\Rightarrow 21x = - 207 \\
\Rightarrow x = - \dfrac{{207}}{{21}} = - \dfrac{{69}}{7} \\
\]
But since $x$ is a digit, its value cannot be negative and fractional. Therefore this case is not possible.
Case 1 - If $z$ is maximum then we have $y = x + 2$ and $z = x + 4$ because they are consecutive odd numbers.
Further it is given that the number is 51 less than 30 times the sum of its digits. We know that the number \[xyz\] is written as $100x + 10y + z$. So from this we have:
\[ \Rightarrow 100x + 10y + z = 30\left( {x + y + z} \right) - 51\]
Putting the values of $y$ and $z$ we’ll get:
\[ \Rightarrow 100x + 10\left( {x + 2} \right) + \left( {x + 4} \right) = 30\left( {x + x + 2 + x + 4} \right) - 51\]
Solving it further for $x$, we’ll get:
\[
\Rightarrow 100x + 10x + 20 + x + 4 = 30\left( {3x + 6} \right) - 51 \\
\Rightarrow 111x + 24 = 90x + 180 - 51 \\
\Rightarrow 21x = 105 \\
\Rightarrow x = 5 \\
\]
Thus the required value of $x$ is 5 and $y = x + 2 = 7,z = x + 4 = 9$
Hence the number \[xyz\] is 579. (B) is the correct option.
Note: Any three digit number \[xyz\] can be written as $100x + 10y + z$. Similarly any two digit number $xy$ can be written as $10x + y$. In the same manner, any $n$ digit number can be written as the sum of digits with the digits multiplied by their respective place value.
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