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For example, if $n = 15^2 \times 5^{18}$ in base $10$. Is there a way to find the upper and lower bounds for the sum of digits of this number? The answer is given to be $6\leq s \leq 140$. Also, is there a way to calculate the exact sum?

Also, how do i find the number of digits of this number?

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    $\begingroup$ Regarding the number of digits of $n$: Note that $\log_{10}n\approx 14.93$. In particular, $14<\log_{10}n<15$, so $10^{14}<n<10^{15}$. This means that $n$ has 15 digits ($n$ is more than $100,000,000,000,000$ but less than $1,000,000,000,000,000$). $\endgroup$ – MPW Oct 10 at 15:58
  • $\begingroup$ My calculator has the exact value: $858,306,884,765,625.$ $\endgroup$ – Thomas Andrews Oct 10 at 16:00
  • $\begingroup$ Well, obviously you can do it exactly by calculating the exact value of it and adding the digits. And clearly whoever came up the is answer didn't. And the ignores that $9|n$ so $s \ge 9$. I don't know what methods the person used to get this range of values but I dont think it matters as the person didn't seem to care bout the exact number. ... oh the number of digits $\approx \log n$ rounded up. $\endgroup$ – fleablood Oct 10 at 16:00
  • $\begingroup$ @ThomasAndrews This question is from an exam where calculators weren't allowed the year this question was given $\endgroup$ – Techie5879 Oct 10 at 16:06
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    $\begingroup$ Well, How precise an answer do they want. You can always say a lower bound is $1$ and an upper bound is a $9$ times the number of digits which most be less than $100^2 *100^{18}=40$ or $360$. It doesn't seem to me that $6\le s \le 140$ is much better than that. It's slightly better but not much better. $\endgroup$ – fleablood Oct 10 at 16:17
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Since the number ends with $5$ , the sum of the digits at the minimum would be $6$ (Since at least one of the other terms is $\ge 1$).

As for the upper bound , As pointed out in the comments , since the number has $15$ digits , I believe the maximum sum should be $9*14 +5 = 131$

Hence the limits should be $6\le s\le 131.$

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    $\begingroup$ But $9|15^2$ so so the sum of the digits is a multiple of $9$ so $9 \le s \le 131$. So the question is how much accuracy does the question want. $\endgroup$ – fleablood Oct 10 at 16:19
  • $\begingroup$ @fleablood considering the answer they provided , I believe they were too lazy to even carry out the answer properly . $\endgroup$ – The Demonix _ Hermit Oct 10 at 16:21
  • $\begingroup$ @fleablood Like you, I saw that the sum of the digits must be divisible by $9$, so I concluded that $9\le s\le 126$ since the numbers between $127$ and $131$ inclusive are not divisible by $9$ $\endgroup$ – Keith Backman Oct 10 at 18:44
  • $\begingroup$ @KeithBackman Just dope slap me..... $\endgroup$ – fleablood Oct 10 at 18:45
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For an upper bound, without logarithms or calculators:

$$\begin{array}{ccc}15^2\cdot5^{18}&<&16^2\cdot 15^{18}\\ &=&16^2\cdot 5^3\cdot5^{15}\\ &<&16^2\cdot 2^7\cdot5^{15}\\ &=&2^{15}\cdot 5^{15}\\ &=&10^{15} \end{array}$$ So the number has at most $15$ digits, one of which is a $5$. So the digit sum can be upper bounded by $9\cdot 14+5=131$.

Whatever method the given answer used seems to have allowed for up to $16$ digits including one $5$.

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Another estimation for the number of digits is that, since $2^{10}> 10^3$, you have that $\log_{10} 2 >0.3$ so $\log_{10} 5 =1-\log_{10} 2 <0.7$ so $\log_{10} 5^{20}=20\log_{10} 5< 14.$

So $5^{20}$ is no more than $14$ digits, and thus $15^2\cdot 5^{18}=9\cdot 5^{20}$ is no more than $15$ digits.

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Well, if you want any upper bound then $n =15^2*5^{18} < 10^{100}$ so it has fewer than a hundred digits each at most $9$ so an upper bound is $900$

And a lower bound is $0$ as the sum is surely positive.

So $0 \le s < 900$

I'm being facetious but whoever did this didn't do much more than that.

I figure s/he figured being an odd multiple of $5$ then last digit was $5$ and having more than one digit the sum $s\ge 6$.

That is bare minimum of work. As $n = 15^2*5^{18}=9*5^{20}$ the number must be a multiple of $9$. So $s \ge 9$.

And the number of digits is $\lfloor \log 9*5^20\rfloor + 1=15$ so the upper bound is $15*9 = 135$. Don't know how the upper bound of $140$ was figured. We know the last digit if $5$; not $9$ so $131$ is an upper bound.

[Credit where credit is due. Even though I knew the last digit it didn't occur to me to consider it an exception to digits being $9$ and thus $131$ is legitimate upper bound. Credit to the Demonix Hermit for that basic observation.]

We can probably figure the first digit of $n=9*5^{20} = 9*10^{20}\frac 1{2^{20}} = 9*\frac 1{1024*1024}10^{20}\approx 9*\frac 1{1000000 + 50000 + 25^2}*10^{20}$ has $8$ and not $9$ as a first digit so $18\le s \le 130$.

Now of course we can actually do the problem. $n =858306884765625$ and $s = 81$.

I'm really unsure what degree of accuracy is expected.

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