
The number of 10-digit numbers such that the product of any two consecutive digits in the numbers is prime number, is
A. 1024
B. 2048
C. 512
D. 64
Answer
593.7k+ views
Hint: To solve this type of problem we have to know the concept like the product of any two consecutive digits is a prime number only when one of them is 1 and the other is a prime number. By using this we are going to solve this problem.
Complete step-by-step solution -
So basically, every other number has to be 1. The other digits have to be 2,3,5 or 7.
Considering the number in the form 1x1x1x1x1x where x = 2,3,5,7.
To place all four digits there are five slots as repetition of numbers is allowed so we can arrange all four numbers in \[4\times 4\times 4\times 4\times 4={{2}^{10}}\] . . . . . . . . . . . (a)
Now considering in another form,
x1x1x1x1x1. Here there are 5 free positions.
To place all four digits there are five slots as repetition of numbers is allowed so we can arrange all four numbers in \[4\times 4\times 4\times 4\times 4={{2}^{10}}\] . . . . . . . . . . . (b)
The total possible numbers are \[{{2}^{10}}+{{2}^{10}}\]
\[2\times {{2}^{10}}\]
\[{{2}^{11}}=2048\]
Thus total possible numbers are 2048.
The answer is option B.
Note: For the ten digit number we can write two forms as seen above. If repetition of numbers is not allowed then we cannot place the same number in the other place if once taken. In the question it is not mentioned anything about repetition so we can assume repetition is allowed.
Complete step-by-step solution -
So basically, every other number has to be 1. The other digits have to be 2,3,5 or 7.
Considering the number in the form 1x1x1x1x1x where x = 2,3,5,7.
To place all four digits there are five slots as repetition of numbers is allowed so we can arrange all four numbers in \[4\times 4\times 4\times 4\times 4={{2}^{10}}\] . . . . . . . . . . . (a)
Now considering in another form,
x1x1x1x1x1. Here there are 5 free positions.
To place all four digits there are five slots as repetition of numbers is allowed so we can arrange all four numbers in \[4\times 4\times 4\times 4\times 4={{2}^{10}}\] . . . . . . . . . . . (b)
The total possible numbers are \[{{2}^{10}}+{{2}^{10}}\]
\[2\times {{2}^{10}}\]
\[{{2}^{11}}=2048\]
Thus total possible numbers are 2048.
The answer is option B.
Note: For the ten digit number we can write two forms as seen above. If repetition of numbers is not allowed then we cannot place the same number in the other place if once taken. In the question it is not mentioned anything about repetition so we can assume repetition is allowed.
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