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# Number Theory – AMC 10A 2016 Problem 22

AMC 10A 2016 Problem 22 solution. See the solution with sequential hints. For some positive integer, the number  has  positive integer divisors

# Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″] For some positive integer $n$, the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$. How many positive integer divisors does the number $81n^4$ have ?

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”3.25″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.0″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.0″]American Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]Number Theory

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0″ open=”off”]8/10

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Elementary Number Theory by David M. Burton

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Check the problem out…give its statement a thorough read. Might appear a bit daunting on the first couple of reads. Think for some time, you could be on to something without any help whatsoever ![/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”4.0″]

Okay, now let’s think about what our first thoughts could be, on the problem. It’s definitely about the n in the problem, which acts as our unknown here.  Can you somehow try finding the n ? Let’s take the first step in that direction. How could we prime factorize 110 ? That’s easy 110 = 2.5.11. Could you take things from hereon to find more about the n ?

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However interestingly the problem says, the number 110. (n^3)  has 110 factors. Just as we saw, 110. (n^3) = 2.5.11.(n^3) Now, let’s use some basic number theoretic knowledge here. How many divisors would 110. (n^3) have then ?  If n=1 Clearly it would have, (1+1). (1+1). (1+1) = 8 divisors.  So see, that’s the idea isn’t it ? Pretty much of plug and play. We actually get to control how many divisors the number has, once we adjust (n^3).  Now you could try some advances…

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Okay, so as we just understood we need to achieve a count of 110 divisors.  If we have 110.(n^3) = 2^(10). 5^(4). 11 which actually conforms to :  (10+1).(4+1).(1+1) = 11.5.2 = 110  So, that implies :   (n^3) = 2^(9). 5^(3), which means, n = 2^(3). 5 Now that we have found out n…the rest dosen’t seem really a big deal. You could do it…try !

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Well, it’s pretty straightforward now.  Let’s call 81.(n^4) equal to some X. First let’s prime factorize 81. That would be 81 = (3^4). So, finally X = (3^4). (2^12). (5^4) How many divisors does that make ? Yes, (4+1).(12+1). (4+1) = 13.25 = 325.

That’s our answer. 325 factors