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
Verified
509.1k+ 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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
10 examples of friction in our daily life
Difference Between Prokaryotic Cells and Eukaryotic Cells
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE
Define least count of vernier callipers How do you class 11 physics CBSE