Which of the following is divisible by the first ten positive integer multiples of 11?
(i) 18810
(ii) 263640
(iii) 3011010	Select the correct option:	A) None	B) ii only	C) i and ii only	D) i and iii only	E) ii and iii only
A positive integer x is divisible by the first five integer multiples of 23 if and only if it is divisible by their least common multiple, 8*9*5*7*11.   8*9*5*7*11 is less than   70*30*10= 
<gadget id="calculator">  70*30*10</gadget>
<output>21_000</output>
21000, (i) cannot possibly be divisible by 8*9*5*7*11.   The integers of 3011010 sum to 6, so 3011010 is divisible by 3, but not 9, so (iii) is not divisible by 8*9*5*7*11.   If (ii) is divisible by 8*9*5*7*11 then 26364 is divisible by 11. So 26364=11y=10y+y for some positive integer y. y must have initial digit 2 and final digit 4. Final and third digit of y must add to 6, so third digit of y is 2. Then similarly, second and third digit add to three, so second digit is 1. But then second and first digit must sum to 6, but first digit is 2, second is 1, so (ii) cannot be divisible by 11.   Nonenswer None
<result>A</result>
Which of the following is divisible by the first ten positive integer multiples of 11?
(i) 18810
(ii) 263640
(iii) 3011010