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06 Dec 2021, 09:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
36% (02:08) correct
64%
(02:17)
wrong
based on 74
sessions
History
Date
Time
Result
Not Attempted Yet
N! is completely divisible by \(13^{52}\). What is the sum of the digits of the smallest such number N?
(A) 9
(B) 11
(C) 15
(D) 16
(E) 19
Are You Up For the Challenge: 700 Level Questions
(A) 9
(B) 11
(C) 15
(D) 16
(E) 19
Are You Up For the Challenge: 700 Level Questions
06 Dec 2021, 13:59
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Between 12 and 157 there are 12 multiples of 13. At 169 there are 2 multiples of 13.
13^14
Between 169 and 326 there are another 12 multiples of 13. At 338 there are 2 multiples of 13.
13^14 * 13^14 =13^28
Between 339 and 495 there again another 12 multiples of 13. At 507 there are 2 multiples of 13.
13^28*13^14 = 13^42
Need 10 more multiples of 13 to reach 13^52.
507+130=637
6+3+7=16,D
Posted from my mobile device
13^14
Between 169 and 326 there are another 12 multiples of 13. At 338 there are 2 multiples of 13.
13^14 * 13^14 =13^28
Between 339 and 495 there again another 12 multiples of 13. At 507 there are 2 multiples of 13.
13^28*13^14 = 13^42
Need 10 more multiples of 13 to reach 13^52.
507+130=637
6+3+7=16,D
Posted from my mobile device
06 Dec 2021, 22:40
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Bunuel
N! = 1*2*3*....*12*13*14*.... *26*27*... 169*170*...
To be divisible by 52 13s, we must get 13 52 times in this product.
13 will give us the first 13 since 13 is prime and we cannot make it using any other smaller factors.
26 will give second.
39 will give third.
...
But 13*13 = 169 will give two 13s.
13*14 will give us fifteenth 13.
...
Similarly, 13*26 will give two 13s
13*27 will give us 29th 13.
...
and 13*39 will give two 13s too.
13*40 will give us 43rd 13
...
so 13*49 will give us 52th 13.
Hence, if N = 13*49 = 637, then 637! will have 52 13s in it.
Sum of digits of 637 = 16
Answer (D)
