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I'm telling you, this question is either sadistic, or I am missing a shortcut of some kind.
X is a 10 digit number that has all the digits from 0 to 9 in it. X's leftmost digit is divisible by 1. The number formed by its leftmost 2 digits is divisible by 2, the number formed by its leftmost 3 is divisible by 3, and so on. Finally, the whole ten digit number is divisible by 10. Which of these is X's sixth digit (counting from the left)?
So far I discovered that there are tons of number sequences up to seven or eight digits that are perfectly peachy this way. Past eight digits... ARGH!
*Bangs head on desk*
No, no, don't tell me what it is. It's no good if you tell me...
Yeah, that question is starting to get on my nerves...The first number can be 1-9, the first two numbers can be 10-98 (all even), the first three digits can be pairs of any number 100-999 (which is divisible by three), the four digits have to be both divisible by two, three and four...
Yup. I don't know if I can give you a hint though without giving the answer away, and I don't want to put it in a spoiler...I actually second guessed whether it worked or not but I went to my calculator and deduced that 41 was a factor of the first three numbers.
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