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# Understand the problem

A sequence of natural numbers is constructed by listing the first $4$, then skipping one, listing the next $5$, skipping $2$, listing $6$, skipping $3$, and, on the $n$th iteration, listing $n+3$ and skipping $n$. The sequence begins $1,2,3,4,6,7,8,9,10,13$. What is the $500,\!000$th number in the sequence ?

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][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, Sequences

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7/10

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”off”]Challenges and Thrills of Pre-College Mathematics

[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”4.0″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” custom_padding=”||153px|25px||”][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″]You could give it a thought first…are you sure you really need a hint ?

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Stuck…? Well, don’t worry. The key to solving this problem is not thinking too much about the skips. We can start with natural numbers, from 1 to 500,000. So, a useful strategy could be to find how many numbers we have actually skipped, n and then add them back accordingly.  So, now could you take things on from here ?

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If you’re a tad bit doubtful of where we’re heading even now, well no problem. Clearly, we can say 999.(1000) / 2   < 500,000 < 1000.(1001) / 2 So, now can you find out how many blocks of gaps we have in the sequence ?

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Now see, finding the blocks of gaps easy ! There’s just one small thing you would have to recall. We began the count from 4…so now, the number of skipped blocks in the sequence = 999 – 3 = 996.  Now to find n, from the number of blocks , we have =  (996.997) / 2 = 496,506 This stands for the number of numbers we skipped. Now concluding this is fairly easy…could you try that out yourself ?

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What remains for us to do is to add back those skipped numbers to the count, 500,000 to obtain the final answer. That gives us = 500000 +496506 = 996506

And we are done !

# Similar Problems

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# 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

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”off”]

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.

# Similar Problems

[/et_pb_text][et_pb_post_slider include_categories=”9″ _builder_version=”3.22.4″][/et_pb_post_slider][et_pb_divider _builder_version=”3.22.4″ background_color=”#0c71c3″][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]
Categories

# 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″]The product $(8)(888\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$. What is $k$?

[/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=”off” _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”]7/10

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”on”]

Elementary Number Theory by David M, Burton

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So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further !

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Think you could do with some help ? Okay let’s get ourselves a headstart. As you might have guessed, sometimes the most fruitful thing to be done is to observe what’s going on in these kind of problems. The intuition is clear, there are too many instances of ‘8‘-s for someone to account for them manually.  So, yes, that’s something we can expect. Try working out with a few simple cases like, k=1, k=2, and so on…

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Okay, so let’s see what’s happening for a few small values of k…
k=1
8 * 8 = 64 k=2
8 * 88 = 704 k=3
8 * 888 = 7104 k=4
8 * 8888 = 71104 …
So, Wait ! Look closely…Do you see something ?

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A really nice pattern is evolving. If you can see it, and yet find it a bit difficult to articulate it mathematically, don’t worry. See for k=4, the product gives us the result 71104. That means, we have the starting digit to be 7. And the ending suffix is 04. What varies are the ones. See, for k=4, we have ( k-2 = 2 ) ones. That’s it ! The generalization should be fairly simple for us to do now… For every k >=2, the product result consists of a 7 to start with, exactly k-2 1’s follow, and we conclude with single occurrences of 0 and 4 each. Now, think…can you take this till the end ?

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Once we’ve seen the pattern, it’s easy to get this done.  What we have noted, we need that to sum up to 1000.  In simple mathematical terms, 7 + (k-2).1 + 0 + 4 = 1000
11 + k = 1002
k = 991

So, we need ‘8’ to come 991 times in the multiplicand, so that the digits sum up to 1000. So, that seals the deal !