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Number Theory Level 2

Find the last digit of ( 1 ! + 2 ! + 3 ! + ⋯ 33 ! ) 33 (1!+2!+3! +\cdots 33!)^{33} .


The answer is 3.

Md Zuhair by Md Zuhair , 20, India

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1 solution

Satwik Murarka
Apr 7, 2017

Solution:

Dividing the expression by 10 will help us to find the last digit,

( 1 ! + 2 ! + 3 ! + 4 ! ⋯ 33 ! ) ≡ 1 + 2 + 4 + 6 + 0 + 0..... + 0 ( m o d 10 ) ( 10 ∣ n ! ≥ 5 ) ( 1 ! + 2 ! + 3 ! + 4 ! ⋯ 33 ! ) 33 ≡ 3 33 ( m o d 10 ) 3 33 ≡ 3 ( m o d 10 ) \begin{aligned}(1!+2!+3!+4!\cdots33!)&\equiv1+2+4+6+0+0.....+0 \pmod{10} \hspace{1cm}(10|n!≥5)\\ (1!+2!+3!+4!\cdots33!)^{33}&\equiv3^{33}\pmod{10}\\ 3^{33}&\equiv3 \pmod{10}\end{aligned}

Last digit of ( 1 ! + 2 ! + 3 ! ⋯ 33 ! ) 33 = 3 (1!+2!+3!\cdots33!)^{33}=\boxed{3}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Beautiful.

Zach Abueg - 4 years, 2 months ago

If you want last two digits then it will be more challenging

Kushal Bose - 4 years, 2 months ago

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Yes, it is. I'm getting an answer of

1 3 33 ≡ ( − 3 ) 11 ( m o d 100 ) ≡ 53 ( m o d 100 ) 13^{33} \equiv (-3)^{11} \pmod{100} \equiv 53 \pmod{100} .

Brian Charlesworth - 4 years, 2 months ago

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Yes I also got this

Kushal Bose - 4 years, 2 months ago

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@Kushal Bose – Since it was your idea, did you want to post the mod 100 version of this problem or shall I?

Brian Charlesworth - 4 years, 2 months ago

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@Brian Charlesworth – U can do it I have no problem

Kushal Bose - 4 years, 2 months ago

Nice solution. We know that 3 33 ≡ 3 ( m o d 10 ) 3^{33} \equiv 3 \pmod{10} because 3 2 ≡ − 1 m o d 10 3^{2} \equiv -1 \mod{10} , and so 3 33 = 3 2 ∗ 16 + 1 ≡ ( ( − 1 ) 16 × 3 ) ( m o d 10 ) ≡ 3 ( m o d 10 ) 3^{33} = 3^{2*16 + 1} \equiv ((-1)^{16} \times 3) \pmod{10} \equiv 3 \pmod{10} .

Brian Charlesworth - 4 years, 2 months ago

Thank you all.@Brian Charlesworth Sir can you tell me how did you compute 1 3 33 ≡ 53 ( m o d 100 ) 13^{33}\equiv53\pmod{100}

Satwik Murarka - 4 years, 2 months ago

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This is probably longer than necessary, but here is my computation.

Since the last two digits of 10 ! 10! are 00 00 we only need to consider

( 1 ! + 2 ! + 3 ! + 4 ! + 5 ! + 6 ! + 7 ! + 8 ! + 9 ! ) ( m o d 100 ) ≡ (1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9!) \pmod{100} \equiv

( 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 ) ( m o d 100 ) ≡ (1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80) \pmod{100} \equiv

213 ( m o d 100 ) ≡ 13 ( m o d 13 ) 213 \pmod{100} \equiv 13 \pmod{13} .

Now 1 3 3 = 13 × 169 ≡ 13 × 69 ( m o d 100 ) ≡ 13^{3} = 13 \times 169 \equiv 13 \times 69 \pmod{100} \equiv

13 ( 70 − 1 ) ( m o d 100 ) ≡ ( 910 − 13 ) ( m o d 100 ) ≡ 13(70 - 1) \pmod{100} \equiv (910 - 13) \pmod{100} \equiv

− 3 ( m o d 13 ) -3 \pmod{13} . So now we have that

1 3 33 = 1 3 3 ∗ 11 ≡ ( − 3 ) 11 ( m o d 100 ) ≡ − 3 ∗ ( 3 5 ) 2 ( m o d 100 ) ≡ 13^{33} = 13^{3*11} \equiv (-3)^{11} \pmod{100} \equiv -3*(3^{5})^{2} \pmod{100} \equiv

− 3 ∗ 4 3 2 ( m o d 100 ) ≡ − 3 ∗ ( 40 + 3 ) 2 ( m o d 100 ) ≡ -3*43^{2} \pmod{100} \equiv -3*(40 + 3)^{2} \pmod{100} \equiv

− 3 ∗ ( 1600 + 240 + 9 ) ( m o d 100 ) ≡ − 3 ∗ 49 ( m o d 100 ) ≡ -3*(1600 + 240 + 9) \pmod{100} \equiv -3*49 \pmod{100} \equiv

3 ∗ 51 ( m o d 100 ) ≡ 53 ( m o d 100 ) 3*51 \pmod{100} \equiv 53 \pmod{100} .

Brian Charlesworth - 4 years, 2 months ago

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