
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
626.7k+ 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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

