Question 1132434
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The least common multiple of 4, 6, and 10 is 60.  So the pattern of which multiples of 4 are or are not multiples of either 6 or 10 will be a cycle of length 60.<br>
The given interval, 25 to 225, is an interval of length 200.<br>
Since the given interval is not an integer multiple of the cycle length, we need to be careful in our counting.<br>
In each cycle of length 60, there are 15 numbers divisible by 4, 5 divisible by 12 (LCM of 4 and 6), 3 divisible by 20 (LCM of 4 and 10), and 1 divisible by 60 (LCM of 4, 6, and 10).  So in each cycle of length 60, the number of integers divisible by 4 but not by 6 or 10 is (15 - (5+3) + 1) = 8.<br>
So in each interval of length 60, there are 8 integers that are multiples of 4 that are not divisible by either 6 or 10, and there are 15-8 = 7 integers that are  multiples of 4 and ARE divisible by either 6 or 10.<br>
The complete interval of 200 is 3 cycles of 60 (28-84, 88-144, and 148-204), plus a partial cycle of 20 (208-224).<br>
3 cycles of 60 gives us 24 integers divisible by 4 and not divisible by either 6 or 10 and 21 integers divisible by 4 but also divisible by either 6 or 10.<br>
Then we need to look at the multiples of 4 from 208 to 224.  2 of them are divisible by either 6 or 10; the other 3 are not.<br>
So in the complete interval from 25 to 225, there are 50 integers divisible by 4; 24+3=27 of them are not divisible by either 6 or 10; 21+2=23 of them are divisible by 6 and/or 10.<br>
So the probability that any multiple of 4 between 25 and 225 is not divisible by either 6 or 10 is 27/50.<br>
Then the probability of choosing two numbers in that interval that are divisible by 4 but not divisible by either 6 or 10 is<br>
{{{(27/50)(26/49) = 351/1225}}}