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# If p,q are distinct and greater than 1, p + q =

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Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3344
Location: India
GPA: 3.12
If p,q are distinct and greater than 1, p + q =  [#permalink]

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02 Mar 2019, 01:26
4
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Difficulty:

95% (hard)

Question Stats:

26% (03:04) correct 74% (02:14) wrong based on 31 sessions

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If $$p,q$$ are distinct and greater than 1, $$p+q =$$ ?

1. $$q^p=p^q$$
2. $$q=p^2$$

Source: Experts Global

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If p,q are distinct and greater than 1, p + q =  [#permalink]

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02 Mar 2019, 02:09
pushpitkc wrote:
If $$p,q$$ are distinct and greater than 1, $$p+q =$$ ?

1. $$q^p=p^q$$
2. $$q=p^2$$

Source: Experts Global

If it isn't entirely clear how to interpret the statements, we'll first try using specific numbers to give us a clue how to proceed (and eliminate ambiguous statements).
This is an Alternative approach.

(1) Trying q = 2 gives 2^p = p^2. So 2^p has to be a square number, and as it is also a power of 2 we'll try 4,16, 64... 4 doesn't work because then p = 2 but p and q are distinct, but 16 could work as then p = 4. If we try 64 we fail again as 8^2 = 64 but 2^8 = 256 and trying any larger number won't work as it makes the exponent larger. So if p = 2 q has to equal 4.
Based on the above, we know that p has to be a power of q, so we can write p = q^n. Substituting gives q^(q^n) = (q^n)^q an using power laws q^(q^n) = q^(nq). Then q^n = nq and since q > 1 we can divide by it to get q^(n-1) = n. So if we pick n = 2 then q = 2 and p = 4 which is what we got above. If we pick n = 3 then q^2 = 3 so q = sqrt(3) and we get some other answer. We won't bother to actually calculate, as this is definitely insufficient.

(2) Once again, we can pick whichever number we like for p, giving another number for q, and then plug this into p + q to get some (different) result.
Insufficient.

Combined:
This is exactly the equation we got to above, with p and q reversed. So we know that n = 2 and therefore there is one solution (q = 4, and p = 2).

If this isn't clear, then simplify the statements by substituting (2) into (1):
(p^2)^p = p^(p^2). Using power laws, this implies that p^(2p)=p^(p^2) and therefore, since p > 1, that 2p = p^2, which gives p = 2 (as p cannot equal 0). Then q = 4 and p + q = 6.
Sufficient.

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If p,q are distinct and greater than 1, p + q =   [#permalink] 02 Mar 2019, 02:09
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