Given-that-N-1-2-3-500-is-the-product-of-the-positive-integers-from-1-to-500-If-N-is-divisible-by-6-k-find-the-largest-possible-value-of-k-

Question Number 115294 by ZiYangLee last updated on 24/Sep/20

Given that N = 1 \times 2 \times 3 \times \dots \times 500 is the

product of the positive integers from 1 to 500 .

If N is divisible by 6^{k}, find the largest possible

value of k .

Answered by Olaf last updated on 25/Sep/20

N = 500!

2, 4, 6, 8 \dots 498, 500 are divisible by 2

(250 numbers)

3, 6, 9, 12 \dots 495, 498 are divisible by 3

3 = 3 \times 1

6 = 3 \times 2

\dots

498 = 3 \times 166

(166 numbers)

250 numbers are divisible by 2

166 numbers are divisible by 3

6^{k} = 2^{k} \times 3^{k} : k_{m a x} = 166

Sorry but i^{'} m wrong .

see mr W .

Answered by mr W last updated on 25/Sep/20

2, 2 \times 2, 3 \times 2, \dots, 250 \times 2 \Rightarrow 2^{250 \times 1}

2^{2}, 2 \times 2^{2}, 3 \times 2^{2}, \dots, 125 \times 2^{2} \Rightarrow 2^{125 \times 2}

2^{3}, 2 \times 2^{3}, 3 \times 2^{3}, \dots, 62 \times 2^{3} \Rightarrow 2^{62 \times 3}

2^{4}, 2 \times 2^{4}, 3 \times 2^{4}, \dots, 31 \times 2^{4} \Rightarrow 2^{31 \times 4}

2^{5}, 2 \times 2^{5}, 3 \times 2^{5}, \dots, 15 \times 2^{5} \Rightarrow 2^{15 \times 5}

2^{6}, 2 \times 2^{6}, 3 \times 2^{6}, \dots, 7 \times 2^{6} \Rightarrow 2^{7 \times 6}

2^{7}, 2 \times 2^{7}, 3 \times 2^{7}, \dots, 3 \times 2^{7} \Rightarrow 2^{3 \times 7}

2^{8} \Rightarrow 2^{1 \times 8}

250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 494

\Rightarrow 500! c o n t a i n s 2^{494}

3, 2 \times 3, 3 \times 3, \dots, 166 \times 3 \Rightarrow 3^{166 \times 1}

3^{2}, 2 \times 3^{2}, 3 \times 3^{2}, \dots, 55 \times 3^{2} \Rightarrow 3^{55 \times 2}

3^{3}, 2 \times 3^{3}, 3 \times 3^{3}, \dots, 18 \times 3^{3} \Rightarrow 3^{18 \times 3}

3^{4}, 2 \times 3^{4}, 3 \times 3^{4}, \dots, 6 \times 3^{4} \Rightarrow 3^{6 \times 4}

3^{5}, 2 \times 3^{5} \Rightarrow 3^{2 \times 5}

166 + 55 + 18 + 6 + 2 = 247

\Rightarrow 500! c o n t a i n s 3^{247}

\Rightarrow 500! c o n t a i n s 2^{494} \times 3^{247}

\Rightarrow 500! c o n t a i n s 6^{247}

\Rightarrow k_{m a x} = 247

Commented by Olaf last updated on 25/Sep/20

Mister W you are wright .

Good job!

Commented by mr W last updated on 25/Sep/20

i t h i n k w e c a n g e n e r a l l y d o l i k e t h i s :

t o g e t k s u c h t h a t n! i s d i v i s i b l e b y

p^{k} w h e r e p = p r i m e,

k = ⌊ \frac{n}{p} ⌋ + ⌊ \frac{n}{p^{2}} ⌋ + ⌊ \frac{n}{p^{3}} ⌋ + \dots

f o r e x a m p l e n = 500, p = 7

k = ⌊ \frac{500}{7} ⌋ + ⌊ \frac{500}{7^{2}} ⌋ + ⌊ \frac{500}{7^{3}} ⌋ + \dots

= 71 + 10 + 1

= 82

\Rightarrow 500! c o n t a i n s 7^{82}

f o r e x a m p l e n = 500, p = 3

k = ⌊ \frac{500}{3} ⌋ + ⌊ \frac{500}{3^{2}} ⌋ + ⌊ \frac{500}{3^{3}} ⌋ + \dots

= 166 + 55 + 18 + 6 + 2

= 247

\Rightarrow 500! c o n t a i n s 3^{247}

Commented by ZiYangLee last updated on 26/Sep/20

Wow!!!

Commented by Lordose last updated on 01/Oct/20

Thanks sir Mr W

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