Solving an Equation on Factorials of Digits | Moscow Mathematical Olympiad 1940

[listen.]

for the case (a55) = a! + 240, you could instantly say that theres no solutions, because for the a > 1 a! is even, but a55 is odd, and a = 1 is not a solution too

[sjoerdo]

An easier way to solve it is to realise that if a, b, c<=5, we have a!+b!+c!<=360, so a<=3. We then find a!+b!+c!<=246, so a<=2. if a=2, we must have 2+b!+c!>=200, so b!+c!>=198. This only has as solution b=c=5, which does not give a solution. so a=1.

we then get 1bc=1+b!+c!. Clearly at least one of b, c must be 5.
If b=5, we get 15c=121+c. this gives no solutions.
If c=5, we get 1b5=121+b, so the last digit of b must be 4. This means b=4, which indeed gives a solution.

[DaveyJonesLocka]

Although I appreciate a rigorous approach to such problems, I think this one could’ve been done with less rigor. I think the following deductive steps lead to an answer fairly quickly. Note: DFS = digit factorial sum of abc.

1. Clearly, all of the digits must be less than seven, due to the three digit restriction.

2. No digit is 6, since this forces the DFS to include at least a 7.

3. There must be at least one 5 to achieve a three digit DFS.

4. Since 3(5!) = 360, the first digit cannot be 5, so there are less than three 5’s.

5. If there were exactly two 5’s, the DFS is bounded between 241 and 264. So if there are two 5’s, the solution must be 255, which fails since its DFS is 244. Hence, there is exactly one 5.

6. Since there is one 5, the DFS is bounded between 122 and 168, forcing the solution to be 15X or 1X5.

7. Since there must be digits of 1 and 5, this bounds the DFS between 122 and 145, forcing the solution to be 125, 135, or 145. Only 145 works.

[bait]

Interesting author typically likes bounding/modulo but instead used alot of case. When u get to <=5 --> <=3*5!=360, starts bounding a. At the end b!%10=4

[harshvadher]

As we restrict a, b, c not to be ≥6 thus a, b, c can't take each 5 and even either both of a, b, c to be 5 so this means a, b, c either is utmost 5 to be occured + rest of digits to be either recurring or distinct.👍
Thus u get finally (1, 4, 5).

[mcwulf]

This is more detective work than number theory! I used a different series of tests, first working through the values of "a" rather like the video which revealed that only a=1 was possible. Then we know if a=1 then 100 <= abc <= 200 so exactly one of b and c must be 5 and the other between 1 and 4 to get the 120 required. This reduces the problem to either:
151 <= 1! + 5! +c! <= 154, for which there are no solutions. Or,
115 <= 1! + b! + 5! <= 145, for which the only solution is b=4.
Hence abc = 145.

[willbishop]

Minor mistake at the beginning, 9! = 362880, not 3628800.

[pcmendes]

In Soviet Russia, the factorials expand you

[tommasobanfi]

why didn't you even analize the case of a=0? there's no solution but i think we should check over it. Btw very nice video thank you

[zzuse]

40585 is the only other non-trivial case