Factorial addition

Algebra Level 2

Let A B C ABC be a 3 digit number, with A A , B B , and C C being distinct nonzero single digits, such that A B C = A ! + B ! + C ! ABC = A! + B! + C! , where x ! x! is the factorial notation ( x ! = x ( x 1 ) ( x 2 ) . . . ( 1 ) ) (x! = x * (x-1) * (x-2)... * (1))

Find A B C ABC . (I.e, if you think the answer is that 678 = 6 ! + 7 ! + 8 ! 678 = 6! + 7! + 8! , you would enter your answer as 678 678 .)


The answer is 145.

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2 solutions

Zach Abueg
Jul 7, 2017

None of the digits can be > 5 > 5 because 7 ! = 5040 > 999 7! = 5040 > 999 and 6 ! = 720 > 700 6! = 720 > 700 . Thus, our digits are bound to [ 1 , 5 ] \left[1, 5\right] . Observe that 4 ! + 3 ! + 2 ! = 32 < 100 4! + 3! + 2! = 32 < 100 , so one of the digits must be 5 5 . Also note that 1 ! + 2 ! + 3 ! + 4 ! + 5 ! = 153 1! + 2! + 3! + 4! + 5! = 153 , meaning that 100 A B C 153 100 \leq \overline{ABC} \leq 153 , so A A must equal 1 1 . Because A , B , C A, B, C are distinct digits, our remaining digit is bound to [ 2 , 4 ] \left[2, 4\right] , making it easier to check for it. Let's examine the remaining cases: 1 B 5 \overline{1B5} or 15 C \overline{15C} .

Case 1: A B C = 15 C \overline{ABC} = \overline{15C}

1 ! + 5 ! + C ! = 15 C 121 + C ! = 150 + C C ! C = 29 \displaystyle \begin{aligned} 1! + 5! + C! & = \overline{15C} \\ 121 + C! & = 150 + C \\ \implies C! - C & = 29 \end{aligned}

There exist no solutions C [ 2 , 4 ] C \in \left[2, 4\right] for the above equation, so we must have A B C = 1 B 5 \overline{ABC} = \overline{1B5} .

Case 2: A B C = 1 B 5 \overline{ABC} = \overline{1B5}

1 + B ! + 5 ! = 1 B 5 121 + B ! = 100 + 10 B + 5 10 B B ! = 16 B = 4 \displaystyle \begin{aligned} 1 + B! + 5! & = \overline{1B5} \\ 121 + B! & = 100 + 10B + 5 \\ \implies 10B - B! & = 16 \\ \implies B & = 4 \end{aligned}

We have A B C = 145 \overline{ABC} = 145 , and verify that, indeed, 1 ! + 4 ! + 5 ! = 145 1! + 4! + 5! = 145 .

Denton Young
Jul 6, 2017

Clearly none of A, B, or C can be 7 or higher, as 7! is a 4 digit number. If 6 were used, 6! =720, so we would need a 7 or higher as well. So the maximum possible is 5. If 5 is not used, the maximum sum is 4! + 3! + 2! = 32, which is not a long enough number. So one of the factorials used must be 5! = 120. Now we need the other two factorials to produce a 5 in the final number. So we need either to reach a total in the 30's (possible only with 4! + 3! = 30, but then 150 doesn't consist of the digits 5, 4, and 3) or the final digit must sum to 5 (possible only with 4! + 1! = 25, so the final total is 120 + 25 = 145.) This works, because 145 = 1! + 4! + 5!

Awesome Question! Is there a mathematical way to solve instead of trial and examining?

Kaushik Chandra - 3 years, 11 months ago

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If you find one, please post it as a separate solution.

Denton Young - 3 years, 11 months ago

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Unfortunately I couldn't find one.

Kaushik Chandra - 3 years, 11 months ago

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