These yelling bases aren't so scary in subscript

Number Theory Level 4

3 3 3 ! + 4 4 4 ! + 5 5 5 ! + … + 9 9 9 ! + X X ‾ X ! \large 33_{3!} + 44_{4!} + 55_{5!} + \ldots + 99_{9!} + \mathrm{\overline { XX } }_{\mathrm{X}!}

Find the remainder (in decimal representation) when the number above is divided by 13 in base 10.

Note: X = 10 \mathrm{X} = 10 .


The answer is 8.

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Jake Lai by Jake Lai , 21, Singapore

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1 solution

Jake Lai
Jun 13, 2015

We first convert the sum into sigma form:

3 3 3 ! + 4 4 4 ! + 5 5 5 ! + … + 9 9 9 ! + X X X ! = ∑ k = 3 10 k k k ! = ∑ k = 1 10 k × k ! + ∑ k = 1 10 k − 8 33_{3!} + 44_{4!} + 55_{5!} + \ldots + 99_{9!} + \mathrm{XX}_{\mathrm{X}!} = \sum_{k=3}^{10} kk_{k!} = \sum_{k=1}^{10} k \times k! + \sum_{k=1}^{10} k - 8

(The 8 8 is from 2 × 2 ! + 1 × 1 ! + 2 + 1 = 8 2 \times 2! + 1 \times 1! + 2 + 1 = 8 .) The second sum is simply ∑ k = 1 10 k = 10 × 11 2 = 55 \displaystyle \sum_{k=1}^{10} k = \frac{10 \times 11}{2} = 55 .

Now, we recall that

∑ k = 1 n k × k ! = ( n + 1 ) ! − 1 ( ∗ ) \sum_{k=1}^{n} k \times k! = (n+1)!-1 \qquad (*)

Hence, we have

∑ k = 1 10 k × k ! + ∑ k = 1 10 k − 8 = 11 ! − 1 + 55 − 8 = 11 ! + 46 \sum_{k=1}^{10} k \times k! + \sum_{k=1}^{10} k - 8 = 11! - 1 + 55 - 8 = 11! + 46

Wilson's theorem states that ( p − 1 ) ! ≡ − 1 ( m o d p ) (p-1)! \equiv -1 \pmod{p} for p p prime. A corollary of this is that ( p − 2 ) ! ≡ 1 ( m o d p ) (p-2)! \equiv 1 \pmod{p} .

Thus, putting this all together we have

11 ! + 46 ≡ ( 13 − 2 ) ! + 7 ≡ 8 ( m o d 13 ) 11! + 46 \equiv (13-2)! + 7 \equiv \boxed{8} \pmod{13}

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Moderator note:

Nice question, but be careful with your calculations.

For completeness's sake, here's a proof of statement ( ∗ ) (*) by induction.

Clearly, the identity is true for n = 1 n = 1 . Now, assume ∑ k = 1 n − 1 k × k ! = n ! − 1 \displaystyle \sum_{k=1}^{n-1} k \times k! = n! - 1 for some n n . Then, we observe that

∑ k = 1 n k × k ! = n × n ! + ∑ k = 1 n − 1 k × k ! = n × n ! + n ! − 1 \sum_{k=1}^{n} k \times k! = n \times n! + \sum_{k=1}^{n-1} k \times k! = n \times n! + n! - 1

= n ! ( n + 1 ) − 1 = ( n + 1 ) ! − 1 = n!(n+1)-1 = (n+1)!-1

This completes our induction.

Jake Lai - 6 years ago

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A direct proof would be to use telescopy.

∑ k = 1 n k × k ! = ∑ k = 1 n ( k + 1 − 1 ) × k ! = ∑ k = 1 n ( ( k + 1 ) ! − k ! ) = ( n + 1 ) ! − 1 ! = ( n + 1 ) ! − 1 \begin{aligned}\sum_{k=1}^n k\times k!&=\sum_{k=1}^n (k+1-1)\times k!\\&=\sum_{k=1}^n \bigg((k+1)!-k!\bigg)\\&=(n+1)!-1!=(n+1)!-1\end{aligned}

Prasun Biswas - 5 years, 11 months ago

your value of ∑ k = 1 10 k = 45 \displaystyle \sum_{k=1}^{10} k = 45 is wrong, it should be 55. Your final answer should be 8.

Pi Han Goh - 6 years ago

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I've been on a careless streak lately... :( Sorry about that.

Jake Lai - 6 years ago

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