Let $b$ be an integer greater than $2$, and let $N_b = 1_b + 2_b + \cdots + 100_b$ (the sum contains all valid base $b$ numbers up to $100_b$). Compute the number of values of $b$ for which the sum of the squares of the base $b$ digits of $N_b$ is at most $512$.

$100_b= 100$ (small low $b$) <<(im not sure what its called)

I honestly am so lost. I don't even know where to begin. Please help.

  • $\begingroup$ Clearly it would help if we could figure out a more workable expression for $N_b$. Have you tried that for a few bases? What did you get? $\endgroup$ – Arthur Mar 6 at 16:38
  • $\begingroup$ I do not know how to do that. So sorry!! $\endgroup$ – Zara Rozia Mar 6 at 16:41
  • $\begingroup$ $N_3$ to $N_5$ are easy to do by hand. Do it (express it in base $b$), and see if you can spot a pattern. For higher bases, google Gauss and the sum of all numbers to $100$. $\endgroup$ – Arthur Mar 6 at 16:43
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    $\begingroup$ How would I know that you struggled with the notation itself? You should've started with that. If you don't even know what the question asks, then the given answer will be of no help to you, and you need a completely different kind of assistance. $\endgroup$ – Arthur Mar 6 at 16:59
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    $\begingroup$ So what you want is for us to answer a problem you don't understand with an answer that you won't understand, in the hopes that studying then in tandem will help you in some way. Have I gotten this right? Well, that is something that I personally won't do. $\endgroup$ – Arthur Mar 6 at 17:35


If $b$ is even, say, $b=2a$ then $N_b=a0a0_b.$

If $b$ is odd, say $b=2c+1$ then $N_b = c(c+1)00_b.$

So you need to find the number of values $a\geq 2$ and $2a^2\leq 512.$

And you need to find the number of values $c\geq 1$ with $2c^2+2c+1\leq 512.$

  • $\begingroup$ I'm not sure where to go from there. Thank you so much for your response!! $\endgroup$ – Zara Rozia Mar 6 at 16:43
  • $\begingroup$ To the proposer: For even $b$ we have $a=b/2$ and $N_b=a0a0_b$ so we require $512\ge a^2+a^2=(b/2)^2+(b/2)^2....$ For odd $b$ we have $c=(b-1)/2$ and $c+1=(b+1)/2$ and $N_b=c(c+1)00_b$ so we require $512\ge c^2+(c+1)^2=((b-1)/2)^2+((b+1)/2)^2.$ $\endgroup$ – DanielWainfleet Mar 7 at 12:16
  • $\begingroup$ I dk why you got a downvote. Looks good to me..........+1 $\endgroup$ – DanielWainfleet Mar 7 at 12:19
  • $\begingroup$ I'm still confused on how we get the answer. $\endgroup$ – Zara Rozia Mar 8 at 21:02
  • $\begingroup$ Nb is the sum of the first k positive integers. What is k? $\endgroup$ – Zara Rozia Mar 8 at 21:04

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