How many 2007-digit numbers exist, in which every two-digit number composed of two sequential digits is divisible either by 17 or 23?
A) 5
B) 6
C 7
D) 9
E) More than 9.
We note that n=1*2*3*….(n-1)*n. If n!=(2^15) *(3^6) *(5^3 )*(7^2 )* 11 * 13, then n=?
A) 13
B) 14
C) 15
D) 16
E) 17
please explain your answer! :)

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TopNewest9 is the answer for the first question Two digit multiples of 17 are 17,34,51,68,85 Two digit multiples of 23 are 23,46,69,92 We can make 2007 digit numbers by 234692.... These sequence can be start from 2,3,4,6 and 9 .Thus giving 5 chances. There is a 4 term extending sequence 8517 which can be merged to 6 (using 68).Thus we will get another 4 numbers which have last digit 8,5,1 and 7 respectively. Consisting of total 9 chances.

For 2nd question ,answer is 16 There is no 17 ,thus it cannot be 17!. 5^3 is present so it must be at least 15!. Checking the power of 2 we can confirm that it is 16! – Muhammed Shemeer K · 4 years, 5 months ago

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From PLK – Zi Song Yeoh · 4 years, 5 months ago

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