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18 Apr 2016, 01:02
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If B and C are digits and 8BC is a 3-digit number that is divisible by 3, which of the following is a possible product of B and C?
A. 1
B. 2
C. 5
D. 6
E. 14
A. 1
B. 2
C. 5
D. 6
E. 14
18 Apr 2016, 03:26
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Bunuel
If B and C are digits and 8BC is a 3-digit number that is divisible by 3, which of the following is a possible product of B and C?
A. 1
B. 2
C. 5
D. 6
E. 14
A. 1
B. 2
C. 5
D. 6
E. 14
8BC is div by 3..
so 8 + B + C should be div by 3...
OR B+C-1 should be div by 3..
1) as 1, 2 and 5 have just two factors 1 and itself, there is ONLY one possiblity of BC, One of them is 1 and other number itself..
Now if any one of B or C is 1, other should be a multiple of 3..
1 = 1*1-- NO
2 = 1*2-- NO
5 = 1*5 -- NO
2) But 6 = 1*6..YES
so if B+C-1 is div by 3.. 1+6-1 = 6, which is div by 3
D
18 Apr 2016, 03:56
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The divisibility rule tells us that a number is only divisible by 3 if the sum of its digits is divisible by 3.
This should ring a bell that 8+B+C can add up to 15 (which is the most logical way for the test takers to go and usually your first point of reference). Now is there an answer choice of which the product numbers can add up to 15?
Yes! 6 is a product of 1*6. 8+1+6=15 and 15 is divisible by 3. Ans D
This should ring a bell that 8+B+C can add up to 15 (which is the most logical way for the test takers to go and usually your first point of reference). Now is there an answer choice of which the product numbers can add up to 15?
Yes! 6 is a product of 1*6. 8+1+6=15 and 15 is divisible by 3. Ans D
