The given expression \[{9^{11}} + {11^9}\] is divisible by
A.7
B.8
C.9
D.10
Answer
Verified
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Hint: Here, we will find the divisibility of the given expression. First , we will find the unit digits for the given terms and add the unit digits of both the terms. Then by using the divisibility rule of integers we will find the divisibility of the expression. A divisibility rule is a method of finding whether the given integer is divisible by an integer without leaving remainder and performing division.
Complete step-by-step answer:
We are given with a term \[{9^{11}} + {11^9}\].
We know that when the number 9 is raised to the odd number of power, then the unit digit will be 9.
For example, \[{9^1} = 9\]; \[{9^3} = 729\]; \[{9^5} = 59049\] and so on.
We know that when the number 9 is raised to the even number of power, then the unit digit will be 1.
For example , \[{9^2} = 81\], \[{9^4} = 6561\] and so on.
Since we are given \[{9^{11}}\], the number 9 is raised to an odd number, so the unit digit will be 9.
We know that when the number 11 is raised to any number of powers, then the unit digit will be 1.
For example \[{11^1} = 11\]; \[{11^2} = 121\]; \[{11^3} = 1331\]; …… and so on
Since we are given \[{11^9}\], the number 11 is raised to an odd number whatever be the power, so the unit digit will be 1.
Now, we will add the unit digits of the given expression.
Thus, we get
\[{9^{11}} + {11^9} = 9 + 1\]
\[ \Rightarrow {9^{11}} + {11^9} = 10\]
Thus, the unit digit of \[{9^{11}} + {11^9}\]is 0.
By Divisibility rule of integers,
We know that if the unit digit is 0, then the number would be divisible by 2, 5, 10.
So, the term \[{9^{11}} + {11^9}\] would be divisible by 2, 5, 10.
Therefore, \[{9^{11}} + {11^9}\] is divisible by 10.
Thus, Option (D) is the correct answer.
Note: We know that it is hard to expand the given term since it has a larger number in the exponents. We can use the binomial expansion for the given expression. We will expand both the expression by using the binomial theorem. The constant term at the binomial expression gets canceled and thus the other binomial coefficients would be divisible by 10. While finding the divisibility it is enough to find the unit digit to find the divisibility rule of integers. For that, we should know the divisibility rule of integers.
Complete step-by-step answer:
We are given with a term \[{9^{11}} + {11^9}\].
We know that when the number 9 is raised to the odd number of power, then the unit digit will be 9.
For example, \[{9^1} = 9\]; \[{9^3} = 729\]; \[{9^5} = 59049\] and so on.
We know that when the number 9 is raised to the even number of power, then the unit digit will be 1.
For example , \[{9^2} = 81\], \[{9^4} = 6561\] and so on.
Since we are given \[{9^{11}}\], the number 9 is raised to an odd number, so the unit digit will be 9.
We know that when the number 11 is raised to any number of powers, then the unit digit will be 1.
For example \[{11^1} = 11\]; \[{11^2} = 121\]; \[{11^3} = 1331\]; …… and so on
Since we are given \[{11^9}\], the number 11 is raised to an odd number whatever be the power, so the unit digit will be 1.
Now, we will add the unit digits of the given expression.
Thus, we get
\[{9^{11}} + {11^9} = 9 + 1\]
\[ \Rightarrow {9^{11}} + {11^9} = 10\]
Thus, the unit digit of \[{9^{11}} + {11^9}\]is 0.
By Divisibility rule of integers,
We know that if the unit digit is 0, then the number would be divisible by 2, 5, 10.
So, the term \[{9^{11}} + {11^9}\] would be divisible by 2, 5, 10.
Therefore, \[{9^{11}} + {11^9}\] is divisible by 10.
Thus, Option (D) is the correct answer.
Note: We know that it is hard to expand the given term since it has a larger number in the exponents. We can use the binomial expansion for the given expression. We will expand both the expression by using the binomial theorem. The constant term at the binomial expression gets canceled and thus the other binomial coefficients would be divisible by 10. While finding the divisibility it is enough to find the unit digit to find the divisibility rule of integers. For that, we should know the divisibility rule of integers.
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