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03 May 2020, 03:45
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EgmatQuantExpert
In the statement
\(Q=2\), in such a case units digit of \(P^{4k+2}\) will be \(9\), which implies that units digit of \(P\) is either \(7\) or \(3\). Hence, units digit of product of \(P\) & \(Q\) would be either \(4\) or \(6\).
Why didn’t you consider the case of 5 ?
5 is also a Prime Number.
ag1991
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11 Jul 2020, 00:32
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Does this qualify as a GMAT question?
30 Nov 2020, 14:47
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EgmatQuantExpert
Given: P and Q are prime numbers less than 70
Target question: What is the units digit of PQ?
Statement 1: Units’ digit of \((P^{4k+2} - Q\)) is equal to 7, where k is a positive integer.
Since the units digit is ODD, I know that one of the values (P or Q) is EVEN and the other term is ODD.
Since 2 is the only EVEN prime, we know that one value is 2
Let's first see if it's possible that P = 2
If k = 1, then 4k+2 = 6
2^6 = 64
So, P^(4k+2) COULD equal 64
We're told that the units digit of \((P^{4k+2} - Q\)) is 7
So, if P^(4k+2) = 64, then Q = 7 would satisfy the condition that the units digit of \((P^{4k+2} - Q\)) is 7
In other words, P = 2 and Q = 7 satisfies statement 1.
PQ = (2)(7) = 14
In this case, the answer to the target question is the unit's digit of PQ is 4
Now let's see if it's possible that Q = 2
If Q = 2, and the digit of \((P^{4k+2} - Q\)) is 7, then the units digit of P^(4k+2) must be 9
If k = 1, then 4k+2 = 6
3^6 = 729
So, P^(4k+2) COULD equal 729
We're told that the units digit of \((P^{4k+2} - Q\)) is 7
So, if P^(4k+2) = 729, then Q = 2 would satisfy the condition that the units digit of \((P^{4k+2} - Q\)) is 7
In other words, P = 3 and Q = 2 satisfies statement 1.
PQ = (3)(2) = 6
In this case, the answer to the target question is the unit's digit of PQ is 6
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: Units digit of the expression \([PQ + Q*(Q+1) - Q^2]\) is a perfect cube
Before we start searching for NEW pairs of values of P and Q that satisfy statement 2, let's first see if either of our values from Statement 1 work.
Let's start with P = 2 and Q = 7
In this case, \(PQ + Q*(Q+1) - Q^2 = (2)(7) + 7(7+1) - 7^2 = 14+56-49=21\)
Since 1 (the units digit) is a perfect cube, P = 2 and Q = 7 satisfies statement 2.
PQ = (2)(7) = 14
In this case, the answer to the target question is the unit's digit of PQ is 4
Now let's try P = 3 and Q = 2
In this case, \(PQ + Q*(Q+1) - Q^2 = (3)(2) + 2(2+1) - 2^2 = 6+6-4=8\)
Since 8 is a perfect cube, P = 3 and Q = 2 satisfies statement 2.
PQ = (3)(2) = 6
In this case, the answer to the target question is the unit's digit of PQ is 6
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
This means the combined statements are NOT SUFFICIENT
Answer: E
Cheers,
Brent
