Total no. of Positive divisers of the form (4n+1))

IIT JEE 2015 Preparation › Algebra questions › Total no. of Positive divisers of the form (4n+1))

\hspace{-16}$Total no. of positive divisers of $\bf{226894500}$ which are is in the form of\\\\ $\bf{(i)\;\;(4n+1)\;\;,}$ Where $\bf{n\in \mathbb{N}}$\\\\ $\bf{(ii)\;\;(4n+2)\;\;,}$ Where $\bf{n\in \mathbb{N}}$\\\\ $\bf{(iii)\;\;(4n+3)\;\;,}$ Where $\bf{n\in \mathbb{N}}$\\\\ $\bf{(iv)\;\;(4n+4)\;\;,}$ Where $\bf{n\in \mathbb{N}}$

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Ketan Chandak ·2013-02-24 05:26:05

Prime factorization of 226894500 is 2²x3³x5³x7⁵

second part and 4th part is pretty easy.For 2nd part,you need to select exactly one power of 2.You are left with 3³x5³x7⁵ which has 4x4x6 factors which is 96.
last part,you have to select both the powers of 2 although the answer remains same.
for 1st and 3rd part you will have to make cases using the conditions that 3,7 which are of the form 4n+3 give a number of the form 4n+3 if their power is odd and of the form 4n+1 otherwise.Vice-versa will be true for 5 which is of the form 4n+1.

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Basic Integral Calculus Operators Symbols Matrix Cases

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Total no. of Positive divisers of the form (4n+1))

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