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Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.

By trial I found $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?

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    $\begingroup$ This problem seems pretty similar. Based off the approach in the video, it may boil down to finding just a few using some logical reasoning about what can and can't happen in this kind of situation. $\endgroup$
    – Decaf-Math
    Apr 12 '20 at 7:55
  • $\begingroup$ Hint: Can $a,b,c$ be all greater than $3$? $\endgroup$ Apr 12 '20 at 8:00
  • $\begingroup$ oh shit i only realised they can't all be greater than 5 @JannikPitt :(( $\endgroup$
    – Gareth Ma
    Apr 12 '20 at 8:12
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Better version.

Notice that $3 \mid x!$ for $x \geq 3$ and $3 \not \mid 2^d$. Therefore, at least one of $a, b, c \leq 2$. WLOG let $a\leq b\leq c$.

If $c\leq 1$, $a\leq b\leq c\leq 1, a!=b!=c!=1$ which gives no solution.

If $c=2$, $a!+b!+2=2^d$. $a,b\in \{0,1\}$ gives $4$ solutions, while $(a,b)=(1,2)$ and $(2,2)$ doesn't give solution.

For the cases below, $c\geq 3$.

If $b\leq 1$, $2+c!=2^d$. Notice that $2^2|c!$ for $c \geq 4$, so $c=3$. This gives $(a,b,c,d)=(a,b,3,3) \forall a,b\in \{0,1\}$ (Extreme laziness)

If $b=2$, $a!+2+c!=2^d$. Note that $a!=1$ doesn't give any solution (parity), so $a=b=2$. $4+c!=2^d$. Noticing that $2^3\mid c!$ for $c \geq 4$, $c=3$. $(a,b,c)=(2,2,3)$ doesn't give a solution.

For the cases below, $c \geq b \geq 3$. $2\mid b!+c!$.

Note that $a!=1$ doesn't give a solution. Therefore, $a=2$.

$2+b!+c!=2^d$. If $c \geq b \geq 4$, $2^3 \mid b!+c!$. Therefore $b=3$ gives $8+c!=2^d$. Note that $c\geq 6$ means $2^4 \mid c!$. Therefore, $c=4$ or $c=5$. Checking shows both of them work.

Therefore all the solutions:

$(a,b,c)=(0,0,2),(0,1,2),(1,1,2),(0,0,3),(0,1,3),(1,1,3),(2,3,4),(2,3,5)$ , up to permutations.

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  • $\begingroup$ Thank you, that's what I was looking for. $\endgroup$
    – Gareth Ma
    Apr 12 '20 at 8:29
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A simplier approach

$a!+b!+c! = 2^d$ where $a,b,c,d€Z$

Notice that $2^d$ must be even, therefore $a!+b!+c!$ must also be even..... We know that the factorial of a number must always be even

Therefore $a!$, $b!$ and $c!$ are all even and $a,b,c > 1$

Since $3*x! ≠ 2^d$, then $a,b,c$ can't be equal, therefore $a < b < c$

$even + even + even = even$

If I sufficiently divide by $2$ it breaks down and at some point becomes

$odd + odd + even = even$

So to find $a$ and $b$, we'll look for two factorials that have a common factor of $2$ or multiples of $2$ and a odd number

$a!$ and $b!$ can be $(2!,3!) = (2×1,2×3)$

It turns out this is the only value that works, because there is no integer that satisfies $x! = 2^n×y$ , where $y$ is odd

$2!+3!+c! = 2^d$

$8+c! = 2^d$

Then the range of values of $c$ is

$c = 4,5,......$

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Without loss of generality $a\le b\le c$, so $a!|2^d\implies a!\in\{1,\,2\}$.

If $a!=1$, $b!+c!$ is odd so $b!=1$ and $c!=2^d-2$, so $c!\nmid4$ and $c\le3$. This gives the solutions $c=2$ and $c=3$.

If $a!=2$, $b!+c!$ isn't a multiple of $4$, so $b\le3$. In particular, if $a=b-2$ then $c!=2^d-4$ is a multiple of $4$ but not $8$ so $4\le c\le7$, and similarly if $a=2,\,b=3$ then $8\le c\le15$. I'll leave you to check these cases.

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Without loss of generality, we can assume $a\leq b\leq c$. Let's do some remarks. I will suppose $d> 3$, the first cases are easy to check by hands and they will coincides with the solutions in $a=1$.

If $a,b,c\geq 3$ then $3$ divides $a!+b!+c!$, but $3$ does not divide $2^d$. Hence $a=1,2$. (the case $a=0$ is equal to case $a=1$; in the solutions you can replace $0$ with $1$)

Case $a=2$

We have $2! + b! + c! = 2^d$ that is $1+\frac{b!}{2}+\frac{c!}{2}=2^{d-1}$.

If $b,c\geq 4$ then the LHS is odd and the RHS is even. Then $b$ has to be $2$ or $3$.

If $b=2$ we have $c!=2^{d}-4 = 4(2^{d-2}-1)$. So we need $c\geq 4$ in order to have a factor $2^2$ in the LHS. But now the LHS have a factor $2^3$ in its factorization, and the RHS don't, a contradiction.

If $b=3$ we have $c!=2^d-8=8(2^{d-3}-1)$. As above we need $c\geq 4$ in order to have a factor $2^3$ in the LHS, but if $c\geq 6$ we have a factor $2^4$ in the LHS factorization and the RHS don't. So $c$ can be only $4$ or $5$.

With these considerations, the solutions are: $$(a,b,c,d) = (2,3,4,5), (2,3,5,7)$$


Case $a=1$

We have $1+b!+c! = 2^d$ that is $b!+c! = 2^d-1$. The RHS is odd, so $b!+c!$ has to be odd. For, we need $b!$ odd and $c!$ even (because $b\leq c$). Hence, the unique case is $b=1$.

So now we have $c! = 2^{d}-2 = 2(2^{d-1}-1)$, and using the same argument used in the case above, we will find that $c$ can be only $2$ or $3$.

With these considerations the solutions are: $$ (a,b,c,d) = (1,1,2,2), (1,1,3,3) $$

Edit: Thanks to Gareth Ma for his remark (case $a=1$).

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  • $\begingroup$ The variables are non-negative, so you need to consider $0$ cases. Also you missed $a=b=1$ case anyways. $\endgroup$
    – Gareth Ma
    Apr 12 '20 at 8:25
  • $\begingroup$ @GarethMa You are right! Thanks! $\endgroup$
    – Menezio
    Apr 12 '20 at 8:39
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Just to give a slightly different approach, let's show that $\max(a,b,c)\le5$, which reduces the problem to a finite search.

Let's assume $a\le b\le c$. As others have noted, we must have $a\le2$, since $a!\mid(a!+b!+c!)$. Now if $b\gt3$, then $4\not\mid(a!+b!)$. It follows that $16\not\mid(a!+b!)$, since $a\le b\le3$ implies $a!+b!\le12\lt16$.

Now suppose $c\ge6$. Then $c!=720n$ for some $n\ge1$ and thus $2^d=a!+b!+c!\gt720$ implies $d\ge10$, in which case

$$a!+b!=2^d-720n=16(2^{d-4}+45n)\implies16\mid(a!+b!)$$

which is a contradiction.

To complete the search, note that if $c=5$ or $4$, then we have $2^d\gt4!=24$, hence $d\ge5$, and thus $8$ divides $2^d-c!=a!+b!$, which occurs if and only if $a!=2$ and $b!=6$ (i.e., $a=2$ and $b=3$), while if $c=3$ or $2$ then $a!+b!=2^d-c!$ is divisible by $2$ but not by $4$, and this occurs if and only if $a!=b!=1$ (i.e, $a,b\in\{0,1\}$). Finally, we cannot have $c=1$ (or $0$) since that would give $a!+b!+c!=1+1+1=3$, which is not a power of $2$. Thus the factorial values $(a!,b!,c!)$ (with $a\le b\le c$) that sum to a power of $2$ are $(1,1,6)$, $(2,6,24)$, and $(2,6,120)$. All other solutions are permutations of these.

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