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The solution posted seems too compilcated...(any number > 5 )! is divisible by 10 , so is modulo 0..that means (n+k) <= 4 and there are only a finite few cases to verify k<=3 && n <= 3.. 4!=4, 3!=6 and 2!=2 ..none of which can equal k

prasadb wrote: The solution posted seems too compilcated...(any number > 5 )! is divisible by 10 , so is modulo 0..that means (n+k) <= 4 and there are only a finite few cases to verify k<=3 && n <= 3.. 4!=4, 3!=6 and 2!=2 ..none of which can equal k The problem says the leftmost digit, not the unit digit.

Suppose that such an integer $n$ exists. for each $1\leq k\leq 9$, there exists an integer $p_{k}$ such that: $k.10^{p_{k}}\leq (n+k)! < (k+1).10^{p_{k}}$ By forming the quotient between two successive inequalities, we find that for each $2\leq k \leq 9$, $10^{p_{k}-p_{k-1}}<n+k<\frac{k+1}{k-1}10^{p_{k}-p_{k-1}}$ (1) As $\frac{k+1}{k-1}<10$ for each $2\leq k \leq 9$, we know that $p_{k}-p_{k-1}$ is a constant (i.e. does not depends on k). Assume $p_{k}-p_{k-1}=l$. Simply use (1) with $k=9$, and you will find: $%Error. "foreach" is a bad command. 1\leq k \leq 9, \ \ 10^{l}\leq n+k<1,25.10^{l}$(2) (by the way, this proves that if such an integer $n$ exists, we have $l\geq 2$ and $n+1\geq 100$) Now, from $(n+1)n!=(n+1)!<2.10^{p_{1}}$ and $(n+4)^{4}n!>(n+4)!\geq 4.10^{p_{4}}$ We get $\frac{(n+4)^{4}}{n+1}>2.10^{p_{4}-p_{1}}=2.10^{3l}$ if $l\geq 3$, then $n+1\geq 1000$ (using (2) ) so $\frac{n+1}{n+4}\geq \frac{1000}{1003}$ so: $(n+4)^{3}=\frac{(n+4)^{4}}{n+1}.\frac{n+1}{n+4}>\frac{1000}{1003}2.10^{3l}$ $n+4>\left(2\frac{1000}{1003}\right)^{1/3}.10^{l}>1,25.10^{l}$ which contradicts (2). Otherwise, $l=2$, and $\frac{(n+4)^{4}}{n+1}>2.10^{6}$ $n+1\geq 100$ (using (2) ) so $\frac{n+1}{n+4}\geq \frac{100}{103}$ so: $(n+4)^{3}=\frac{(n+4)^{4}}{n+1}.\frac{n+1}{n+4}>\frac{100}{103}2.10^{6}$ $n+4>\left(2\frac{100}{103}\right)^{1/3}.10^{2}>123$ so: $n+9>125$ which again contradicts (2).

Xell wrote: By forming the quotient between two successive inequalities I don't think we can do this. a>b and c>d does not mean a/c > b/d. Take 4>3 and 100>1

dgreenb801 wrote: I don't think we can do this. a>b and c>d does not mean a/c > b/d. Take 4>3 and 100>1 But he takes a/d > b/c. It's still valid since $ a,b,c,d>0$.

for every natural number $ m$,denote $ N(m)$ as: $ N(m) = \frac {m}{10^{d(m) - 1}}$. in which $ d(m)$ is the number of digits of $ m$.now notice that $ 1\leq N(m) < 10$.now it can be easily shown that for every two natural numbers $ m,\ell$: $ N(m\ell)\leq N(m)\cdot N(\ell)$ (1) now let $ n$ be a natural number such that for $ k = 0,1,2,\ldots ,9$,the leftmost digit of $ (n + k)!$ equals $ k$.if $ 2\leq k\leq 9$ then there exist a nonnegative integer $ r$,and a real number $ a$,for which $ (n + k)! = a\times 10^r$ and also $ k < a < k + 1$.with a similar argument there exist a real number $ b$ and a nonnegative integer $ s$,for which $ (n + k - 1)! = b\times 10^s$ such that $ k - 1 < b < k$.hence: $ 1 < N(n + k) = N\left(\frac {(n + k)!}{(n + k - 1)!}\right) = \frac ab < \frac {k + 1}{k - 1}\leq 3$ (2) on the other hand the inequality $ N(m + 1)\leq N(m)$ holds only if $ 9\leq N(m)$ and also it follows from (2) that: $ 1 < N(n + 2) < N(n + 3) < \cdots < N(n + 9)\leq\frac 54$ and using (1) we get that: $ N\left((n + 2)!\right)\leq N\left((n + 1)!\right)N(n + 2) < 20\times\frac 54$ $ N\left((n + 3)!\right)\leq N\left((n + 2)!\right) N(n + 3) < 20\times\left(\frac 54\right)^2$ $ N\left((n + 4)!\right)\leq N\left((n + 3)!\right)N(n + 4) < 20\times\left(\frac 54\right)^3 < 4$ but this contradicts the assumption that the leftmost digit of $ (n + 4)!$ equals $ 4$. QED

Here's another approach. Let $ \color[rgb]{.35,.35,.35} n \geq 4$ be a counterexample to the theorem. Logarithms are base $ \color[rgb]{.35,.35,.35} 10.$ Let $ \{n\}$ denote the fractional part of $ \log n,$ so that $ \{n\}$ increases until $ \log n$ jumps an integer, at which point $ \{n\}$ decreases close to $ 0,$ starts increasing again, and so on. Prior to a jump, $ \{n\}$ is at least $ \tfrac 12,$ as $ \log n$ increments by at most $ \tfrac 2n.$ Divide $ [0,1]$ at the points $ \log 2, \ldots, \log 9,$ and see how $ \{(n+k)!\}$ belongs to the $ k-$th segment, for $ 1 \leq k \leq 9.$ No jump occurs in $ \{n+2\}, \ldots, \{n+9\},$ as the lengths of the segments force them all to be $ < \tfrac 12,$ so this sequence is increasing. But $ \log 2 \geq \small size \{n+6\} + \cdots + \{n+9\} \geq \{n+2\} + \cdots + \{n+5\} \geq \log \tfrac 52$ is a contradiction.

Define $s(x)$ as \[s(x) = \frac{x}{10^{\lfloor \log(x)\rfloor }}\] Note that we can bound \[(n+9) = \frac{(n+9)!}{(n+8)!} < \frac{10}{8}\] We now note that \[\frac{(n+4)!}{(n+1)!} < (n+9)^3 < \frac{10}{8}^3 < 2\]However, we also have that \[\frac{s((n+4)!)}{s((n+1)!)} > \frac{4}{2}=2\]This gives us $2< \frac{s((n+4)!)}{s((n+1)!)}\leq \frac{(n+4)!}{(n+1)!} < 2 $, which is absurd.

orl wrote: Prove that there is no positive integer $n$ such that, for $k = 1,2,\ldots,9$, the leftmost digit (in decimal notation) of $(n+k)!$ equals $k$. Isn't $n=4 , k=1$ a counterexample to this statement? @below Oops sorry got the question wrong

for $k = 1, 2, 3, \dots, 9$...

This problem has a lot of different valid approaches.

We proceed by contradiction. Let the reduced copy of a positive integer $n$ be $a$, where $n=a\cdot 10^b$ for a nonnegative integer $b$ and a terminating decimal $a$ such that $1\le a<10$. If the leading digit of $n$ is $k$ so that $k$ is from $1$ to $9$, then the reduced copy of $n$ is between $k$ and $k+1$. Since the reduced copy of $(n+1)!$ is between $1$ and $2$, and the reduced copy of $(n+2)!$ is between $2$ and $3$, the reduced copy of $n+2$ must be between $\frac{2}{2}=1$ and $\frac{3}{1}=3$. Doing this again, the reduced copy of $n+3$ must be between $\frac{3}{3}=1$ and $\frac{4}{2}=2$. The reduced copy of $n+9$ must be between $\frac{9}{9}=1$ and $\frac{10}{8}=1.25$. Note that the lower bound is always $1$ and the upper bound starts at $3$ and slowly reduces to $1.25$. Claim: The reduced copy of all numbers from $n+2$ to $n+9$ now must be between $1$ and $1.25$. Proof. Suppose not, because the numbers from $n+2$ to $n+9$ are consecutive, then there will be a number from $n+2$ to $n+9$ with reduced copy between $9$ and $10$ which is obviously impossible because that does not fit into any of the earlier ranges (that had a lower bound of $1$ and an upper bound of at most $3$). $\blacksquare$ Note that the reduced copy of $(n+1)!$ is between $1$ and $2$. Since the reduced copies of $n+2, n+3, n+4$ are all between $1$ and $1.25$, the reduced copy of $(n+4)!$ is less than $2(1.25)^3=3.90625<4$, contradiction as it must be between $4$ and $5$.

ISL Marabot solve Suppose for the sake of contradiction there does. Let \[f(x)=\frac{x}{10^{\lfloor \log_{10} x\rfloor }}.\] Note $1\le f((n+1)!)<2$ and $2\le f((n+2)!)<3$. Since $\frac{(n+2)!}{(n+1)!}=n+2$, we note $\frac{2}{2}=1<f(n+2)<\frac{3}{1}=3$. Similarly, we can find the bounds $1\le f(n+k)< \frac{k+1}{k-1}\le 3$ for $k=2,3,4\ldots,9$ and $1\le f(n+9)<\frac{10}{8}=1.25$. Claim: For all $k=2,3,4,\ldots,9$, $1<f(n+k)<1.25$. Proof: Suppose not. Then since $n+2,n+3,\ldots,n+9$ are all consecutive, at least one of $\{f(n+2),f(n+3),\ldots,f(n+8)\}$ is between $9$ and $10$, a contradiction as that doesn't fit into the bounds we had found earlier. $\blacksquare$ Now we bound $f((n+1)(n+2)(n+3))$. Note $4\le f((n+4)!)<5$ and $1\le f((n+1)!)<2$. So $\frac{4}{2}=2<f((n+1)(n+2)(n+3))<\frac{5}{1}=5$. This is a contradiction as $f((n+1)(n+2)(n+3))=f(n+1)f(n+2)f(n+3)\le (1.25)^3=1.953125<2$

Just an Error

Xell wrote: Suppose that such an integer $n$ exists. for each $1\leq k\leq 9$, there exists an integer $p_{k}$ such that: $k.10^{p_{k}}\leq (n+k)! < (k+1).10^{p_{k}}$ By forming the quotient between two successive inequalities, we find that for each $2\leq k \leq 9$, $10^{p_{k}-p_{k-1}}<n+k<\frac{k+1}{k-1}10^{p_{k}-p_{k-1}}$ (1) As $\frac{k+1}{k-1}<10$ for each $2\leq k \leq 9$, we know that $p_{k}-p_{k-1}$ is a constant (i.e. does not depends on k). Assume $p_{k}-p_{k-1}=l$. Simply use (1) with $k=9$, and you will find: $%Error. "foreach" is a bad command. 1\leq k \leq 9, \ \ 10^{l}\leq n+k<1,25.10^{l}$(2) (by the way, this proves that if such an integer $n$ exists, we have $l\geq 2$ and $n+1\geq 100$) Now, from $(n+1)n!=(n+1)!<2.10^{p_{1}}$ and $(n+4)^{4}n!>(n+4)!\geq 4.10^{p_{4}}$ We get $\frac{(n+4)^{4}}{n+1}>2.10^{p_{4}-p_{1}}=2.10^{3l}$ if $l\geq 3$, then $n+1\geq 1000$ (using (2) ) so $\frac{n+1}{n+4}\geq \frac{1000}{1003}$ so: $(n+4)^{3}=\frac{(n+4)^{4}}{n+1}.\frac{n+1}{n+4}>\frac{1000}{1003}2.10^{3l}$ $n+4>\left(2\frac{1000}{1003}\right)^{1/3}.10^{l}>1,25.10^{l}$ which contradicts (2). Otherwise, $l=2$, and $\frac{(n+4)^{4}}{n+1}>2.10^{6}$ $n+1\geq 100$ (using (2) ) so $\frac{n+1}{n+4}\geq \frac{100}{103}$ so: $(n+4)^{3}=\frac{(n+4)^{4}}{n+1}.\frac{n+1}{n+4}>\frac{100}{103}2.10^{6}$ $n+4>\left(2\frac{100}{103}\right)^{1/3}.10^{2}>123$ so: $n+9>125$ which again contradicts (2). can someone help me ? why $p_{k}-p_{k-1}$ is a constant Edited: please

Let $f(n)$ be equal to $\frac{n}{10^{\lfloor\log_{10}{n}\rfloor}}.$ In other words, $f(n)$ is the result you get when you divide $n$ by $10$ until you get a number in the interval $[1,10).$ Note that $f$ is multiplicative, that is, $f(mn)=f(m)f(n).$ Assume for the sake of contradiction that $\lfloor f((n+k)!)\rfloor = k$ for $k=1,2,\dots,9.$ Note that $2\le f((n+2)!)< 3$ and $1\le f((n+1)!)< 2$ implies $1<f(n+2)<3.$ We also have from the same logic, $1<f(n+9)<\frac{5}{4}.$ Suppose $10^i<n+9<\frac{5}{4}\cdot 10^i$ for nonnegative integer $i.$ If $i=0$ then $n+9<1.25$ which doesn't make sense. If $n+9<12.5$ then $n<3.5.$ $n=1,2,3$ immediately fail the test that $(n+1)!$ begins with a $1.$ Thus, $i\ge 2$ so we can assume that $n+2,n+3,\dots,n+9$ have the same number of digits. Otherwise, $n+2$ will not be in the range $(10^i,3\cdot10^i)$ This implies that $f(n+2)<f(n+3)<f(n+4)<f(n+9)<\frac{5}{4}.$ Thus, $100\cdot 10^i<n+2<125\cdot 10^i$ for some $i$ and $100\cdot 10^i<n+3<125\cdot 10^i.$ Note that $1\le f((n+1)!)<2$ so $1<f((n+2)!)< 2.5.$ Finally, we have $1<f((n+4)!)<2\cdot 1.25^3=3.90625$ which is a contradiction!

Assume towards a contradiction that such $n$ exists. Then, for some $x, y\in \mathbb{Z}^+$, $$8\cdot 10^x \leq (n+8)!<9\cdot 10^x$$$$9\cdot 10^y \leq (n+9)\cdot (n+8)!<10^{y+1}$$$$10^{y-x}<n+9<\frac54\cdot 10^{y-x}$$By similar logic, for some $a$, $$10^a<n+2<2\cdot 10^a$$Also, $$10^{y-x}-7< n+2<\frac54\cdot 10^{y-x}-7$$The only way these can both be satisfied is if $$10^{y-x}< n+2<\frac54\cdot 10^{y-x}-7$$ So, $$10^{y-x}<n+1, n+2, \ldots, n+9<\frac54\cdot 10^{y-x}$$For some $b\in \mathbb{Z}^+$, $$10^b\leq(n+1)!<2\cdot 10^b$$$$10^{b+3(y-x)}<(n+4)!<2\cdot (\frac54)^3\cdot10^{b+3(y-x)}$$$$10^{b+3(y-x)}<(n+4)!<\frac{125}{32}\cdot10^{b+3(y-x)}<4\cdot10^{b+3(y-x)}$$So, the first digit of $(n+4)!$ can't be $4$, a contradiction.

We first note that: $$2\cdot10^{k_1}<\frac{(n+4)!}{(n+1)!}<5\cdot10^{k_1}$$$$1.4\cdot10^{k_2}<\frac{(n+7)!}{(n+4)!}<2\cdot10^{k_2}$$It suffices to show that for large enough $n$, $$1<\frac{(n+5)(n+6)(n+7)}{(n+2)(n+3)(n+4)}<\frac{14}{5},$$and all $k$ below that large enough $n$ don't satisfy the wanted. The lower bound is obvious, and the upper bound happens when $n\ge6$: \begin{align*} &\frac{(n+5)(n+6)(n+7)}{(n+2)(n+3)(n+4)} \\ =&\frac{n+5}{n+2}\cdot\frac{n+6}{n+3}\cdot\frac{n+7}{n+4} \\ =&\left(1+\frac{3}{n+2}\right)\cdot\left(1+\frac{3}{n+3}\right)\cdot\left(1+\frac{3}{n+4}\right) \\ <&\left(1+\frac{3}{8}\right)^3<(1.4)^3<\frac{14}{5}. \end{align*}It can be easily verified that $n=1,2,3,4,5$ don't work, so we are done. QED.

For $1\leq k\leq 9$, let $$(n+k)!= a_k\textbf{e}b_k$$in scientific notation. We have that $k\leq a_k<k+1$ for all $1\leq k\leq 9$. Claim: $b_k$ is an arithmetic sequence. The idea here is that the mantissa doesn't "overflow" as we multiply here. So if at any point the arithmetic progression breaks, we would either have a multiplier that is about 10 times the previous multiplier, which obviously can't happen, or we would have a multiplier that is less than the previous multiplier (which also doesn't happen due to factorial). Let the common difference be $d$. Then going from the first term to the third term, the multiplier is at least $(3/2)10^{2d}$. Going from the seventh term to the ninth term, the multiplier is at most $(10/7)10^{2d}$, which is smaller, contradiction since the multiplier is supposed to increase.

Solved with GeNuAlCo_pi FTSOC, let's assume that the question statement is wrong and there exists such $n$. Now, we get, $10\cdot 10^{x+y} > (n+9)! \geq 9\cdot 10^{x+y}$ and, $9\cdot 10^{x} > (n+8)! \geq 8\cdot 10^{x}$ for some $(x,y) \in \mathbb N^2$. Now by dividing these two inequalities, we get $\frac{10}8 \cdot 10^y > n+9 > 10^y$. So, $n+9$ has first digit $1$. So, $n+1$ must have first digit $1$ or $9$. But, $3\cdot 10^{x+y} > (n+2)! \geq 2\cdot 10^{x+y}$ and, $2\cdot 10^{x} > (n+1)! \geq 10^{x}$ for some $(x,y) \in \mathbb N^2$ Then again dividing gives, $3 \cdot 10^y > n+2 > 10^y\implies 3 \cdot 10^y > n+1 \geq 10^y$ Therefore, $n+1$ also has first digit $1$. So, $n+1, n+2, \dots, n+9$ all have same number of digits. let the number of digits be $b+1$. Then, $$\frac{10}8 \cdot 10^b > n+9 > n+1 \geq 10^b$$$$\implies 1.25 > \frac{n+9}{10^b} > \frac{n+1}{10^b} \geq 1.00$$From this we get, $$1.953125 = (1.25)^3 > \frac{(n+4)(n+3)(n+2)}{10^{3b}} > (1.00)^3 = 1$$Therefore, first digit of $(n+4)(n+3)(n+2)$ is $1$. But using the previous trick, $5\cdot 10^{x+y} > (n+4)! \geq 4\cdot 10^{x+y}$ and, $2\cdot 10^{x} > (n+1)! \geq 10^{x}$ for some $(x,y) \in \mathbb N^2$. And then, $$ 5\cdot 10^y > (n+4)(n+3)(n+2) > 2\cdot 10^y$$So we get, first digit of $(n+4)(n+3)(n+2)$ is $2, 3$ or $4$. But this is a contradiction. So our assumption was wrong and the problem statement is true. $\blacksquare$

We note that $\{ \log_{10}((n+k)!) \} \in [\log_{10} k , \log_{10} (k+1)]$. Then $\{log_{10} ( (n+k+1)(n+k) ) \} \in \left[ \log_{10} \left(\frac{n+k+1}{n+k} \right), \log_{10} \left(\frac{n+k+2}{n+k+1}\right) \right]$ by subtraction of $n+k+2$ and $n+k$. This works for $1 \le k \le 7$. Now note $\log_{10} \left(\frac{n+3}{n+2}\right) \le \{ \log_{10} ((n+2)(n+1)) \} \le \log_{10} \left(\frac{n+4}{n+1} \right)$ and $\log_{10} \left(\frac{n+10}{n+7} \right) \le \{ \log_{10} ((n+9)(n+8)) \}$. Thus $\{ \log_{10} ((n+2)(n+1)) \} \ge \{ \log_{10} ((n+9)(n+8))\}$. Since $\log_{10} ((n+9)(n+8)) > \log_{10} ((n+2)(n+1))$, we have $\lfloor \log_{10} ((n+9)(n+8)) \rfloor > \lfloor \log_{10} ((n+2)(n+1)) \rfloor$. Thus \[ \log_{10}((n+9)(n+8)) - \log_{10} ((n+2)(n+1)) > 1 -\log_{10} \left( \frac{n+4}{n+1} \right) + \log_{10} \left( \frac{n+9}{n+8} \right) \implies \frac{(n+9)(n+8)}{(n+2)(n+1)} > 10 \cdot \frac{(n+9)(n+1)}{(n+4)(n+8)} \implies \frac{n+8}{n+2} \cdot \frac{n+8}{n+1} \cdot \frac{n+4}{n+1} > 10 \implies n \le 5. \]We conclude by taking $k=6-n$ and noting $7 \neq 6-n$.

Surely the most overcomplicated solution here?

Assume this is possible. Let $f(x)$ be $x \cdot \frac{1}{10^ {\lfloor \log_{10} x \rfloor}}$. Now note that $f(f(x_1)f(x_2) \cdots f(x_i)) = f(x_1 x_2 \cdot x_i)$, as both sides are just $x_1 x_2 \cdot x_i$ multiplied by a power of $10$, and since both values are at least $1$ and less than $10$ the power of $10$ must be the same. Note that we have $i \le f((n + i) !) < i + 1, i + 1 \le f((n + i + 1)!) < i + 2$ for $1 \le i \le 8$, so we can write $f((n + i + 1)!) = f(f((n + i)!)f(n + i + 1))$. We see that the term inside the $f$ on the right side must be less than $10$, otherwise it must be at least $10(i + 1)$, which is impossible since $f((n + i)!)$ is less than $i + 1$, and $f(n +i + 1)$ is less than $10$. So we can right $f(n + 1 + i) = \frac{f((n + 1 + i)!)}{f((n + i)!)}$, so the bounds $1 < f(n + 1 + i) <1 + \frac 2i$ can be establish. Now we note $f(x) > f(x + 1)$ implies a transition in the number of digits, so $f(x) \ge 9$, but this is impossible for $x = n + i$ by the bounds, so we can conclude that $f(n + 2) < f(n + 3) < \cdots < f(n + 9)$. Now note that we can use previous arguments to get $6 > f((n +2 )(n + 3)(n + 4)(n + 5)) > \frac 52$. Now using the bounds on each $f(n + i)$ we see that $f(n + 2)f(n + 3)f(n + 4)f(n + 5) < 15$, but since $f$ of that is between $2.5$ and $6$, we know that $2.5 < f(n + 2)f(n + 3)f(n + 4)f(n + 5) < 6$. Now use the left bound, we have $2.5 < f(n + 2)f(n + 3)f(n + 4)f(n + 5) < f(n + 9)^4 < (\frac 54)^4$, which forces $\frac 52 4^4 < 5^4$ or $256 < 250$, contradiction.