IMO B
N = a2b, where N is a multiple of 3 and a multiple of 5
so b must be 5 or 0
1)a+b is an even integer
when b=5, a=5
when b=0, a =4
not sufficient
2)a/b=1
in this case b not equal to 0. so b must be 5 and hence a must be 5
N is 525
sufficient
the three digit pos
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tohellandback
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Definitely "B".
Since "5" has less multiples available in 3 digit integer, lets choose "5" to test. 5 has multiple integers that end with either 0 or 5. With the statement 2 showing first and third digit in the 3-digit integer being the same, once you test (121, 222, 323, 424, 525, 626, 727, 828 and 929), only 525 has both factors of 3 and 5.
Since "5" has less multiples available in 3 digit integer, lets choose "5" to test. 5 has multiple integers that end with either 0 or 5. With the statement 2 showing first and third digit in the 3-digit integer being the same, once you test (121, 222, 323, 424, 525, 626, 727, 828 and 929), only 525 has both factors of 3 and 5.












