\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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1 Answers
1057Prime 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.
