Sum of factorials of digits
Let be the unique three digit integer satisfying . Find .
Clarification: denotes the concatenation of the digits , not multiplication.
The answer is 145.
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N = ABC = A! + B! + C!. That means the number should be of three digits. now, 6! = 720, but 7! is of four digits. So the possibility of A is 1,2,3,4,5,6 and so of B and so of C. Now we consider numbers from 100 to 199. So we get, A=1, so now we will Consider those B and C which satisfies 98 < B! + C! <199, This shows {B,C} = {(4,5),(5,4)}.
Similarly when we consider numbers from 200 to 299, we get A=2, Then we have to consider those B and C which satisfies 195 < B! + C! < 296, This shows B and C should be equals to 5. i.e. {B,C} = {(5,5)}.
BUT, when we consider numbers, 300-399 (i.e. A=3), 400-499 (i.e. A=4), 500-599 (i.e. A=5), we get no value for B and C. Now, when when A=6, that means the number is between 600 and 699. but A! = 6! = 720 > 699. So when A is 6 there is no solution.
So ultimately, the solution may possibly be {A,B,C} = {(1,4,5),(1,5,4),(2,5,5)} now, 1! + 4! +5! = 145 is true, but the others are not true. So the solution is A=1, B=4, C=5. And it is unique.