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Divisibility question which deals with the units digit. (There are some that deal with the tens digit as well) So we will only look at the units digits of the question. Both 373 and 423 end with a 3. Thus list out all the units digit that occur with different powers of 3: 3^1 = 3 3^2 = 9 3^3= 27 3^4= 81 3^5= 243 we stop here because we can see that after this the units digit is repeating (it becomes 3 again, same as in 3^1).
Now on to the statements. 1. says that p^2 + 2pq + q^2 = 36 Thus (p+q)^2 = 36 and thus p+q = +6/-6. It says in the question that p and q are positive integers So p + q cannot be -6
Thus possible values of p and q are p q 1 5 : Units digit of 3^1 =3 and 3^5 =3. thus Units digit of question stem will be 3 x 3= 9 2 4 : Same as above, units digit will be 9 and 1 respectively and thus 9 x 1 = 9 3 3 : 7 and 7 respectively, and 7 x 7 = 49 and thus again units digit is 9 4 2 : No need to calculate for this as it will be same as 2,4 5 1 : No need to calculate
Thus we can see for given values of p and q the answer is always the same i.e 9. so 1. is sufficient
2. Given that only p = 2, q can take any value. Thus it is clearly insufficient
The answer is thus A
The main trick in the question was to realise that even after multiplying two different numbers we are getting the same answer of 9. Most people after finding that there can be multiple values of p and q would have deemed 1. as insufficient and jumped to the conclusion that you need both 1. and 2. to solve the question and thus get the wrong answer as C. Identifying tricks that the gmat uses is essential for scoring 700+
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