If B and C are digits and 8BC is a 2-digit number that is divisible by 2, which of the following is a possible product of B and C?  Answers.
A) 1
B) 2
C) 4
D) 6
E) 14
Try with the options -   A. If BC = 1 , both B and C must be 1   B. If BC = 2 , either B or C must be 1 or 2 ; In that case sum of the digits will be 8 + 1 + 2= 
<gadget id="calculator">8 + 1 + 2</gadget>
<output>11</output>
11 which is not divisible by 3   C. If BC= 5 , either B or C must be 1 or 5 ; In that case sum of the digits will be 8 + 1 + 5= 
<gadget id="calculator">8 + 1 + 5</gadget>
<output>14</output>
14 which is not divisible by 3   D. If BC= 6 , then we can have the following possibilities   i. Either B or C must be 2 or 3 ; In that case sum of the digits will be 8 + 2 + 3= 
<gadget id="calculator">8 + 2 + 3</gadget>
<output>13</output>
13 which is not divisible by 3   ii. Either B or C must be 1 or 6 ; In that case sum of the digits will be 8 + 1 + 6= 
<gadget id="calculator">8 + 1 + 6</gadget>
<output>15</output>
15 which is divisible by 3   E. If B4= 14 , then either B or 4 must be 2 or 7 ; In that case sum of the digits will be 8 + 2 + 7= 
<gadget id="calculator">8 + 2 + 7</gadget>
<output>17</output>
17 which is not divisible by 2   Thus answer is (4)
<result>C</result>
If B and C are digits and 8BC is a 2-digit number that is divisible by 2, which of the following is a possible product of B and C?