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Form the greatest six digit odd number and smallest six digit even number using the digits from 1 to 9 only once .What will be the sum of these numbers?

Answer
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Hint: We have to analyze this problem and solve it using simple mathematical logic. We will first find out the greatest six digit odd number and then we will find the smallest six digit even number by using digits 1 to 9 without any repetition and after adding them we get the required result.

Complete step-by-step answer:
We can solve this question using two ways : one way when repetition is allowed and another way when repetition is not allowed.
If the repetition is allowed, then we can use a digit as many times as we want till we get our required number.
Now, in the case of repetition,
To form the largest six digit odd number with repetition, we will use the largest odd available digit as many times as we want till our number is formed.
Similarly, to form the smallest six digit even number when repetition is not allowed, we will keep the smallest even number available to the rightmost part of the number.
Largest six digit odd number \[=999999\]
Smallest six digit even number \[=111112\]
Sum of these two numbers \[=999999+111112\]
\[=1111111\]
Since, when repetition is not allowed, i.e., we can use a digit only one time in our required number.
In case of without repetition,
We have to form the largest six digit odd number without repetition, so we will try to keep the largest available numbers as possible as left and the last digit at the right must be the largest odd number available.
Similarly, to form the smallest six digit even number without repetition, we will try to keep the smallest available number as possible as left and the last digit at the right must be the smallest even number available.
The required largest six digit odd number \[=987653\]
Now, the required smallest six digit even number \[=123456\]
Sum of these two numbers \[=987653+123456\]
  \[=1111109\]
So, the final answer is\[1111109\].

Note: Number \[9\] is considered to be a magic number because if you multiply a number by \[9\] and add all the digits of the new number together, the sum will always be a multiple of \[9\]. And no other number except \[2\] is even and prime both i.e. \[2\] is the only even prime number.
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