If the given expression \[n\in N\],\[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\] then is divisible by which one of the following?
a)1904
b)2000
c)2002
d)2006
Answer
Verified359.7k+ views
Hint: To solve the question, we have to apply the formula that \[{{a}^{n}}-{{b}^{n}}\] is divisible by (a – b). Apply the formula to all terms of the expression to find common divisible factors of the expression.
Complete step-by-step answer:
We know that \[{{a}^{n}}-{{b}^{n}}\] is divisible by (a – b). By applying the formula we get
\[{{121}^{n}}-{{25}^{n}}\] is divisible by (121 - 25) = 96
\[{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]is divisible by (1900 – (-4)) = 1900 + 4 = 1904
We know \[96=16\times 6,1904=16\times 119\]
Thus, the common factor of 96, 1904 is 16.
Thus, 16 can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
By applying the above formula for another set of terms of expression, we get
\[{{121}^{n}}-{{\left( -4 \right)}^{n}}\]is divisible by (121 – (-4)) = 121 + 4 = 125
\[{{1900}^{n}}-{{25}^{n}}\]is divisible by (1900 - 25) = 1875
We know \[1875=15\times 125\]
Thus, the common factor of 125, 1875 is 125.
Thus, 125 can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
Thus, we get both 16 and 125can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
This implies that the product of 16 and 125 can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
We know that product of 16 and 125 = \[16\times 125=2000\]
Thus, 2000 can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
Hence, option (b) is the right answer.
Note: The possibility of mistake can be interpreted that 1904 divides the given expression because it divides \[{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]. But it is not divisible by the other part of the expression, only common factors can divide the whole expression. The alternative to solve the questions is equal to substitute n = 1 in the given expression, the calculated value is equal to 2000. Hence, the other options can be eliminated.
Complete step-by-step answer:
We know that \[{{a}^{n}}-{{b}^{n}}\] is divisible by (a – b). By applying the formula we get
\[{{121}^{n}}-{{25}^{n}}\] is divisible by (121 - 25) = 96
\[{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]is divisible by (1900 – (-4)) = 1900 + 4 = 1904
We know \[96=16\times 6,1904=16\times 119\]
Thus, the common factor of 96, 1904 is 16.
Thus, 16 can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
By applying the above formula for another set of terms of expression, we get
\[{{121}^{n}}-{{\left( -4 \right)}^{n}}\]is divisible by (121 – (-4)) = 121 + 4 = 125
\[{{1900}^{n}}-{{25}^{n}}\]is divisible by (1900 - 25) = 1875
We know \[1875=15\times 125\]
Thus, the common factor of 125, 1875 is 125.
Thus, 125 can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
Thus, we get both 16 and 125can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
This implies that the product of 16 and 125 can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
We know that product of 16 and 125 = \[16\times 125=2000\]
Thus, 2000 can divide the expression \[{{121}^{n}}-{{25}^{n}}+{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]
Hence, option (b) is the right answer.
Note: The possibility of mistake can be interpreted that 1904 divides the given expression because it divides \[{{1900}^{n}}-{{\left( -4 \right)}^{n}}\]. But it is not divisible by the other part of the expression, only common factors can divide the whole expression. The alternative to solve the questions is equal to substitute n = 1 in the given expression, the calculated value is equal to 2000. Hence, the other options can be eliminated.
Last updated date: 26th Sep 2023
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