Factorial Sums

We need to find all k such that $2^k = a! + b! + c! + d!$ WLOG, let $a \ge b \ge c \ge d$ . Note that $d! | a!, d! | b!, d! | c!$ , hence $d! | 2^k$ . We can see that d=1 or 2: if $d \ge 3$ , then since $d! | 2^k$ , $2^k$ must contain a factor of 3, clearly impossible.

If $d=1$ , then $c$ must also be 1, as all $2^k$ are even for positive k. If $b \ge 4$ , then $4| b!, a!$ , so $a! + b! + c! + d! \equiv 2 \pmod {4}$ , clearly impossible for $k>1$ . We can use casework:

If b=1, then a must be 1 as well. Thus, $2^k =4$ .
If b=2, then $2^k = a! +4$ , so a<4. Quick testing yields no solutions.
If b=3, then $2^k = a! +8$ , so a<6. Quick testing yields $2^k = 128, 32$ .

Now, we try $d=2$ . Clearly, $c$ can only be 2 or 3.
If $c=2$ , then we need $2^k = a! + b! + 4$ . Since $b<4$ (by divisibility), $b$ can only be 2 or 3. Similarly, $a$ can only be 2 or 3 as well. We can quickly check that (a,b,c,d)=(3,3,2,2) or (2,2,2,2) work, so $2^k = 16, 8$ .
If $c=3$ , we need $2^k = a! +b! + 8$ . In this case, $b$ can only be 4 or 5. Therefore, we get $2^k = a! + 32$ (impossible for $a!$ to have exactly 5 factors of 2) or $2^k = a! +128$ . In the latter case, a can be 8 or 9. Neither works.

Our final answer is $128 + 32 + 16 + 8 + 4 = 188$ .

Let's say that $2^k=a!+b!+c!+d!$ where $a\geq b\geq c\geq d$ . Note that $k\geq 2$ since otherwise at least one of $a,b,c,d$ has to be zero. Since $d$ is the smallest number, all factorials of numbers greater than it divide $d!$ . For $d\geq 3, 3|d!$ , and by extension $3$ divides the right side of the equation, but not the left one(since $2^k$ only contains $2$ s). Hence, we have two cases: $d=1$ and $d=2$ .

Case 1 : $d=1$ , we now have $2^k-1=a!+b!+c!$ . Since all factorials of numbers greater than one are even, and $2^k-1$ is odd, we have that $c=1$ , since $c$ is now the smallest of the three numbers. The equation now becomes $2^k-2=a!+b!$ . If $b>3$ we have that $4|a!+b!$ , but $4\not | 2^k-2$ . Hence, $b\in\{1,2,3\}$ .

For $b=1$ we have $2^k-3=a!$ . For $a>2, 3|a!$ but $3\not |2^k-3$ , so $a\in\{1,2\}$ . For $a=1$ we have that $\boxed{k=2}$ and for $a=2$ we have no solutions.

For $b=2$ we have $2^k-4=a!$ . If $a<4, 2|a!$ but $4\not |a!$ (hence impossible since $4|2^k-4$ and if $a\geq 4$ we have that $8|a!$ but $8\not |2^k-4$ . Hence, this case produces no solutions.

For $b=3$ we have $2^k-8=a!$ . Using the logic from the previous part we can bound $a$ as follows: $4\leq a\leq 5$ . By checking, we see that both cases give solutions: $\boxed{k=5}$ and $\boxed{k=7}$ respectively.

Case 2: $d=2$ . The equation transforms to $2^k-2=a!+b!+c!$ . Following a logic similar to the one for the previous case, we see that $c\in\{2,3\}$ . Note: $c\neq 1$ since $c\geq d$ .

For $c=2$ we have $2^k-4=a!+b!$ . Similarly to the previous case we get that $b\in\{2,3\}$ .

For $b=2$ , considering divisibility by $3$ we get that the only solution is $a=2\implies \boxed{k=3}$ .

For $b=3$ , the left side is divisible by $2$ only, while the right one is divisible by $4$ for $a\geq 4$ . $a=2$ is not a solution, while $a=3$ gives us $\boxed{k=4}$ .

The only case left to consider now is $c=3$ . The equation now is $2^k-8=a!+b!$ . Checking divisibility by $8$ we see that $3\leq b\leq 5$ .

For $b=3$ we have $2^k-14=a!$ , which checking divisibility by $4$ doesn't give solutions.

For $b=4$ we have $2^k-32=a!$ . If $a\leq 6$ , $32|2^k-32$ while $32\not |a!$ . If $a>6$ , $64|a!$ while $64\not |2^k-32$ . Hence, no solutions.

For $b=5$ we have $2^k-128=a!$ . Using a similar logic to the previous one, we bound $a$ as follows: $8\leq a\leq 9$ . However, by checking manually, we see that neither case gives us solutions.

Hence, all solutions we found are $k=2,3,4,5,7$ and the sum is $2^2+2^3+2^4+2^5+2^7=4+8+16+32+128=188$ .

If 2^n is expressible as a sum of four factorials, then, since all factorials except for 1! and 2! are divisible by 3, we would necessarily have 1's and/or 2's in the sum; and since all factorials except for 1! are even, we would have to have either four 1's, two 1's or none; so finally 2^n would have to be expressible in one of the following forms: 1+1+1+1 1+1+2+2 1 +1 + 2 + a! 1 +1 + a! +b! 2+2+2+2 2 + 2 + 2+ a! 2 + 2 +a! +b! 2 + a! +b! +c!

AS 3 <= a <= b <= c.

The first case yields 4, which is a power of 2. The second case yields 6, which is not.

In the third case, a!+4 must be a power of 2; this is not true for a=3, nor is it the case for any higher value for a, since in that case a! is divisible by 8 and therefore a!+4 must be divisible by 4 and by no higher power of 2.

In the fourth case, a must be =3, since otherwise both a! and b! are divisible by 8 and 2+a!+b! is congruent to 2 modulo 8. So b!=2^n-8 is divisible by 8 and by no higher power of 2, this means that b is at most 5, since 6!=720=16*45. And indeed, if b is 4 or 5, 1+1+3!+b! is 32 or 128.

The fifth case yields 8, a power of 2.

In the sixth case we want a!+6 to be a power of 2 , which is impossible, since it is divisible by 3 for a>=3.

In the seventh case we want a!+b!+4 to be a power of 2. If a and b are both >=4, a!+b!+4 is congruent to 4 mod 8; so a must be =3 and we want b!+10 to be a power of 2. That is the case for b=3, but not for b=4, nor for b>=5, since in that case b!+10 is divisible by 5.

In the eighth case, a must be =3, since otherwise a!+b!+c!+2 is congruent to 2 modulo 8. So we want b!+c!+8 to be a power of 2. b=c=3 doesn't work, so b! and c! must both be divisible by 8, i.e. b and c are both >=4. But they can't both be >=6, since in that case b! and c! are both divisible by 16 and b!+c!+8 is only divisible by 8. So b is either 4 or 5, and we want 32+c! or 128+c! to be a power of 2. Since 8!, hence also c! for any c>=8, is divisible by 128, c!+32 is congruent to 32 mod 128 and cannot be a power of 2. Likewise, as 10! is divisible by 256, c!+128 is congruent to 128 mod 256 and cannot be a power of 2 for c>9. So we only have to check a few potential cases, which yield no further solutions.

Thus, the only powers of 2 that can be written as a sum of four factorials are 4, 8, 16, 32, and 128 4+8+16+32+128=188.

at least 1 of the 4 factorials must be smaller than 3, if not the number itself will be a multiple of 3. so we try 1!+1!+1!+1!=4,2!+2!+2!+2!=8,3!+3!+2!+2!=16,4!+3!+1!+1!=32, and 5!+3!+1!+1!=128, since 64, 256 and 512(because 4x5!<520) doesn't work, and the other powers will make the answer exceed 1000, we stop at 128 and the answer is 4+8+16+32+120=188.

First, notice that at least one of the factorials must be $1$ or $2$ , otherwise the sum is divisible by $3$ .

Some of the small powers of $2$ can be represented as sums of factorials as follows:

$4=1!+1!+1!+1!,\ 8=2!+2!+2!+2!,\ 16=2!+2!+3!+3!,\ 32=1!+1!+3!+4!$

The following table shows the order of prime $2$ in $n!$ .

$\begin{array}{ccccccccccc} n & 1&2&3&4&5&6&7&8&9&10\\ ord_2n! &0&1&1&3&3&4&4&7&7&8 \end{array}$ From now on, we suppose that the power of $2$ is at least $2^6$ . Denote the number of times that $n!$ appears in the sum by $a_n$ . We already know that $a_1+a_2\geq 1.$ Because $2^n$ is even, $a_1$ must be even. If it is $4$ , then we get $4=1+1+1+1$ . If it is $2$ , we get $2^m=1+1+k!+l!$ for some $k$ and $l$ . Considering this sum modulo $4$ , we conclude that $a_2+a_3$ is odd, so either $a_2=1,\ a_3=0$ or $a_2=0, \ a_3=1$ .

In the first case, we get $2^m=4+k!$ . Since we assumed that $m\geq 6$ , the order of $2$ in $k!$ must be $2$ , which is impossible. In the second case, we get $2^m=8+k!$ . Here the order of $2$ in $k!$ must be $3$ , so $k$ equals $4$ or $5$ . In the first case the sum is $32=2^5$ , that we already considered, and in the second case we get $128=1!+1!+3!+5!$ , so $128=2^7$ has to be counted.

We are left with the case $a_1=0.$ Considering the sum modulo $4$ , we get that $a_2+a_3$ is even. Because $a_2\geq 1$ , and $a_2+a_3\leq 4$ , we get the following cases:

Case 1. $a_2=4,\ a_3 =0$ . Then the sum is $8=2^3$ .

Case 2. $a_2=3,\ a_3 =1$ . Then the sum is $12$ , which is not a power of $2$ .

Case 3. $a_2=2,\ a_3=2$ . Then the sum is $16=2^4$ .

Case 4. $a_2=2,\ a_3 =0$ . Then $2^m=4+k!+l!$ , where $k$ and $l$ are at least $4$ . This is impossible, because the right hand side is $4$ modulo $8$ .

Case 5. $a_2=1,\ a_3=3$ . Then the sum is $20$ , which is not a power of $2$ .

Case 6. $a_2=1, \ a_3=1$ . Then $2^m=8+k!+l!$ . Because the left hand side is divisible by $16$ , we conclude that $a_4+a_5$ is odd. If $a_4=1, \ a_5=0$ , we get $2^m=32 + l!,$ with $l\geq 6$ . Because $m\geq 6$ , the order of $2$ in $l!$ must be $5$ , which is impossible. If $a_4=0,\ a_5=1$ , then $2^m=128+ l!$ with $l\geq 6.$ If $l\geq 10,$ then the right hand side is $128$ modulo $2^8$ , so $2^m\leq 128$ , which is impossible. So the only cases possible are $l=7,8,9$ . By inspection, they do not produce powers of $2$ .

Putting it all together, we get $2 + 3 + 4 + 5 + 7 = 21$ .

1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 8! = 41320

Smallest sum of 4 factorial = 4 x 1! = 4

2^0 = 1 (2 < 4) 2^1 = 2 (2 < 4) 2^2 = 4 (1! + 1! + 1! + 1! = 4) 2^3 = 8 (2! + 2! + 2! + 2! = 8) 2^4 = 16 (2! + 2! + 3! + 3! = 16) 2^5 = 32 (4! + 3! + 1! + 1! = 32) 2^6 = 64 (no possible combination) 2^7 = 128 (5! + 3! + 1! + 1! = 128) 2^8 = 256 (no possible combination) 2^9 = 512 (no possible combination) 2^10 = 1024 (no possible combination) 2^11 = 2048 (no possible combination)

Assume that there are no more sums of 4 factorials that can be expressed as powers of 2.

4 + 8 + 16 + 32 + 128 = 188

Final ans: 188

Clearly the power of 2 will not exceed 9 as answer limit is 0-999. Hence checking for these nine numbers is very easy. This will result into final solution as 4 + 8 + 16 + 32 + 128 = 188. Another concept which I can be a tentative solution is the factoradic notation . For more details refer, here

I listed all the factorials from 1 to 7, any number higher than that will have intervals that are too great to be able to add up to any power of two. I then calculated all the powers of two and tested if I could add up to them using four factorials. Again, you don't need to go any higher than a certain number of powers because the intervals become too great again.

2 cannot be expressed as a sum of 4 factorials,2^2 = 1!+1!+1!+1!,2^3=2!+2!+2!+2!,2^4 = 3!+3!+2!+2!,2^5 = 4!+3!+1!+1!,2^7 = 5! + 3! +1!+1! and other values cannot be expressed as the sum of the last digits of the powers of 2 matches only the values of 2!,1!,3!,4!,5!

Begin by writing all powers of $2$ from $2$ to $1024$ ). Also, write all factorials from $1$ to $120$ .

Powers of $2$ : $\{2,4,8,16,32,64,128,256,512\}$ . Factorials: $\{1,2,6,24,120,720\}$ .

Now, individually test each power of $2$ . $2$ cannot be represented like this. The following powers of $2$ can be written as the sum of these four factorials: $4 - \{1,1,1,1\}$ , $8 - \{2,2,2,2\}$ , $16 - \{6,6,2,2\}$ , and $128 - \{120,6,1,1\}$ .

$64$ cannot be expressed as the sums because the highest factorial less than $64$ is $24$ . $64$ would have to be comprised of as many $24$ s as possible, $\{24,24\}$ . With a high leftover amount of $16$ , $16$ cannot be written as the sum of two factorials. Thus, $64$ cannot.

The same logic is used for $256$ and $512$ . $256$ would be $\{120,120\}$ with a remainder of $16$ . $512$ would be $\{120,120,120,120\}$ , which doesn't even come to $512$ . Thus, only the above numbers work.

It can easily be shown that no numbers past $128$ can be written as a sum of four factorials, but for time's sake, we only have to test up to $512$ because the maximum input for brilliant.org is 999.

$4+8+16+32+128=188$ .

several powers of 2 come which when added give188