chetan2u wrote:
Are the digits A and B consecutive integers?
(1) A*B=72.
(2) \(15\leq{A+B}\leq{19}\), where A+B is a prime number.
When you recognize the fact that A and B are
digits and that
digits can only range from 0 to 9, it becomes much easier:
(1) A*B = 72
The only single-digit factor-pairs of 72 are 8*9 and 9*8.
If you don't recognize it straight away, you could write down all possible pairs (1*72, 2*36,...) and will realize it at some point.
Both possibilities are cons. integers.
Every other pair has at least one double-digit factor (e.g. 6 * 12), which makes it impossible for both A and B to be digits.
Since the only possible solutions are cons. integers
SUFF(2) \(15\leq{A+B}\leq{19}\), where A+B is a prime number.
The only two primes within this range are 17 and 19
Let's how we can construct them through a sum:
19: since 9 is the highest possible digit (and if either A or B is 9), you would need to add 10, which is not a digit
Therefore, 19 is not an option
A+B needs to be 17. Let's check the possibilites, starting with A being the highest possible digit:
If A = 9, B = 8
If A = 8, B = 9
If A = 7, B = 10
You see again that the only possible digits are 9 & 8.
Every solution under/above this combination need at least a double-digit integer (>= 10)
Therefore, the only possible solution are cons. integers
SUFFAnswer D