Proof & Number Theory

Part (i). We want all pairs of positive integers $(n,p)$ with $p$ prime satisfying $n!+5=p$.

For $n\ge 5$, the factorial $n!$ contains $5$ as a factor, so $n!+5\equiv 0(\mathrm{mod}5)$, meaning $p$ would be divisible by $5$. The only prime divisible by $5$ is $5$ itself, but $n!+5\ge 5!+5=125>5$, so there are no solutions for $n\ge 5$.

We check $n=1,2,3,4$ directly:$$n=1:1!+5=6=2\times 3\text{(not prime)}$$$$n=2:2!+5=7\text{(prime)}✓$$$$n=3:3!+5=11\text{(prime)}✓$$$$n=4:4!+5=29\text{(prime)}✓$$The solutions are $(n,p)\in \{(2,7),(3,11),(4,29)\}$.

Part (ii). We seek all pairs of positive integers $(n,m)$ satisfying$$1!\times 3!\times \cdots \times (2n-1)!=m!$$Small cases. Check $n=1,2,3,4,5,6$ directly.

- $n=1$: $1!=1=1!$, giving $(1,1)$.
- $n=2$: $1!\times 3!=6=3!$, giving $(2,3)$.
- $n=3$: $1!\times 3!\times 5!=6\times 120=720=6!$, giving $(3,6)$.
- $n=4$: $1!\times 3!\times 5!\times 7!=720\times 5040$. Note $5040=7!$ and $720=6!$, so the product equals $6!\times 7!=6!\times 7\times 6!=7\times (6!{)}^2$. More directly: $720\times 5040=3628800=10!$, giving $(4,10)$.
- $n=5$: Multiply by $9!=362880$. The LHS has exactly two factors of $7$ (one from $7!$), but $14!$ is needed to supply two factors of $7$, yet no factor of $11$ appears anywhere in $1!\times 3!\times \cdots \times 9!$, whereas any $m!\ge 11!$ contains $11$. So no solution.
- $n=6$: Multiply further by $11!$. Now two factors of $7$ force $m\ge 14$, but the LHS contains no factor of $13$, while $14!$ does. No solution.

Large cases ($n\ge 7$). By Theorem 1, the LHS exceeds $(4n)!$, so $m>4n$. By Theorem 2, there exists a prime $q$ with $2n<q<4n$. Since $q<4n<m$, the prime $q$ divides $m!$. However, $q>2n-1$, so $q$ cannot divide any factor $(2k-1)!$ for $k\le n$ (since $2k-1\le 2n-1<q$), meaning $q$ does not divide the LHS. This is a contradiction, so there are no solutions for $n\ge 7$.

The complete set of solutions is $(n,m)\in \{(1,1),(2,3),(3,6),(4,10)\}$.

Part (i).

Write $x={\sum }_{r=0}^{n-1}{a}_r\times 10^r$ and $d(x)={\sum }_{r=0}^{n-1}{a}_r$. Subtracting:$$x-d(x)=\sum _{r=0}^{n-1}{a}_r(10^r-1)$$Since $a_r\ge 0$ and $10^r-1\ge 0$ for all $r\ge 0$, every term is non-negative, so $x-d(x)\ge 0$.

For divisibility by 9, factor: $10^r-1=(10-1)(10^{r-1}+10^{r-2}+\cdots +1)=9\times (\text{integer})$, so each term $a_r(10^r-1)$ is divisible by 9, and hence so is their sum. Thus $9\mid x-d(x)$.

Part (ii).

Divisibility claim. Write $x-44d(x)=44(x-d(x))-43x$. By part (i), $9\mid x-d(x)$, so $9\mid 44(x-d(x))$. Hence $9\mid x-44d(x)$ if and only if $9\mid 43x$. Since $\gcd (43,9)=1$, this holds if and only if $9\mid x$.

Bounding $n$. Suppose $x=44d(x)$. Each digit is at most 9, so $d(x)\le 9n$, giving $x\le 44\times 9n=396n$. Since $x$ has $n$ digits and its leading digit is positive, $x\ge 10^{n-1}$. So we need $10^{n-1}\le 396n$.

For $n=4$: $10^3=1000\le 1584$. For $n=5$: $10^4=10000>1980=396\times 5$. So $n\le 4$.

Finding $x$. Since $x=44d(x)$, we have $9\mid x-44d(x)=0$, so $9\mid x$. Also $44\mid x$ (directly). Since $\gcd (9,44)=1$, we need $396\mid x$, so $x=396k$ for some positive integer $k$.

With $n\le 4$, the four-digit bound $x\le 396\times 4=1584$ gives candidates $x\in \{396,792,1188,1584\}$. Note all have digit sum 18.

Checking: $44\times 18=792$. So:
- $x=396$: $44d(396)=792\ne 396$
- $x=792$: $44d(792)=44\times 18=792$ ✓
- $x=1188$: $44d(1188)=792\ne 1188$
- $x=1584$: $44d(1584)=792\ne 1584$

The unique solution is $x=792$.

Part (iii).

Suppose $x=107d(d(x))$. Since $d(d(x))\le d(x)\le 9n$, we get $x\le 107\times 9n=963n$. Combined with $x\ge 10^{n-1}$, we need $10^{n-1}\le 963n$. For $n=5$: $10000>4815$, so $n\le 4$.

To show $9\mid x$: write$$x-107d(d(x))=107(x-d(x))+107(d(x)-d(d(x)))-106x$$By part (i) applied twice, $9\mid x-d(x)$ and $9\mid d(x)-d(d(x))$, so the first two terms are divisible by 9. If $x=107d(d(x))$ then the left side is zero, hence $9\mid 106x$. Since $\gcd (106,9)=1$ (as $106=2\times 53$), we get $9\mid x$.

Also $107\mid x$ directly. Since 107 is prime and $\gcd (107,9)=1$, we need $963\mid x$, giving $x=963k$. With $x\le 963\times 4=3852$, the candidates are $x\in \{963,1926,2889,3852\}$.

For all four, $d(x)\in \{18,18,27,18\}$ respectively, so $d(d(x))=9$ in every case, and $107\times 9=963$.

Checking:
- $x=963$: $107d(d(963))=963$ ✓
- $x=1926$: $107d(d(1926))=963\ne 1926$
- $x=2889$: $107d(d(2889))=963\ne 2889$
- $x=3852$: $107d(d(3852))=963\ne 3852$

The unique solution is $x=963$.

Part (i).

The first block of $n+k$ consecutive integers starts at $c$ and ends at $c+n+k-1$. Using the formula $\frac{\text{(number of terms)}}{2}(\text{first}+\text{last})$, its sum is$$\frac{n+k}{2}(2c+n+k-1).$$The next block of $n$ consecutive integers starts at $c+n+k$ and ends at $c+2n+k-1$, giving sum$$\frac{n}{2}(2c+3n+2k-1).$$Setting these equal and expanding both sides:$$(n+k)(2c+n+k-1)=n(2c+3n+2k-1).$$Expanding the left side: $2nc+n^2+nk-n+2ck+nk+k^2-k$. Expanding the right side: $2nc+3n^2+2nk-n$. Cancelling $2nc$, $2nk$, and $-n$ from both sides leaves$$n^2+2ck+k^2-k=3n^2,$$which rearranges to $2n^2+k=2ck+k^2$, as required. Every step is an equivalence, so the "if and only if" holds. $\square$

Part (ii).

Case (a): $k=1$. The equation becomes $2n^2+1=2c+1$, so $c=n^2$. This holds for every positive integer $n$, with $c=n^2$.

Case (b): $k=2$. The equation becomes $2n^2+2=4c+4$, i.e. $n^2=2c+1$. For $c$ to be a positive integer, $n^2$ must be odd, so $n$ must be odd. Any odd integer $n\ge 3$ works, with $c=\frac{n^2-1}{2}$. (Check: $n=1$ gives $c=0$, not a positive integer, so $n\ge 3$.)

Part (iii): $k=4$.

The equation gives $2n^2+4=8c+16$, so $n^2=4c+6=4(c+1)+2$. Thus $n^2\equiv 2(\mathrm{mod}4)$. But any integer $n$ satisfies $n^2\equiv 0$ or $1(\mathrm{mod}4)$ (since if $n$ is even, $n^2\equiv 0$; if $n$ is odd, $n^2\equiv 1$). A remainder of $2$ is impossible, so there are no solutions. $\square$

Part (iv): $c=1$.

Part (iv)(a). Setting $c=1$ in $2n^2+k=2k+k^2$ gives $2n^2=k(k+1)$. We need $k(k+1)$ to be twice a perfect square. Computing $k(k+1)$ for small $k$: $2,6,12,20,30,42,56,72,\dots$ The values that are twice a perfect square are $2=2\times 1^2$ and $72=2\times 6^2$. This gives two solutions: $(k,n)=(1,1)$ and $(k,n)=(8,6)$.

Part (iv)(b). Suppose $(K,N)$ is a solution with $c=1$, so $2N^2=K(K+1)$. We must show that $N^{\prime }=3N+2K+1$ and $K^{\prime }=4N+3K+1$ also satisfy $2N^{\prime 2}=K^{\prime }(K^{\prime }+1)$.

It suffices to show $2N^{\prime 2}-K^{\prime }(K^{\prime }+1)=0$. Expanding:$$2N^{\prime 2}=2(3N+2K+1{)}^2=18N^2+8K^2+24NK+12N+8K+2,$$$$K^{\prime }(K^{\prime }+1)=(4N+3K+1)(4N+3K+2)=16N^2+9K^2+24NK+12N+9K+2.$$Subtracting:$$2N^{\prime 2}-K^{\prime }(K^{\prime }+1)=2N^2-K^2-K=2N^2-K(K+1)=0,$$using the hypothesis. Hence $(K^{\prime },N^{\prime })$ is also a solution. $\square$

Part (iv)(c). Starting from $(K,N)=(8,6)$ and applying the recurrence once:$$N^{\prime }=3(6)+2(8)+1=35,K^{\prime }=4(6)+3(8)+1=49.$$So $(k,n)=(49,35)$ is a solution. Applying the recurrence again:$$N^{\prime \prime }=3(35)+2(49)+1=204,K^{\prime \prime }=4(35)+3(49)+1=288.$$So $(k,n)=(288,204)$ is another solution. The two further values of $k$ are $49$ (with $n=35$) and $288$ (with $n=204$).