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For all numbers p and q (where p ≠ 0 ), the operation $ is defined

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For all numbers p and q (where p ≠ 0 ), the operation $ is defined  [#permalink]

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New post 10 Apr 2019, 11:12
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Question Stats:

25% (01:20) correct 75% (01:00) wrong based on 12 sessions

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For all numbers p and q (where p ≠ 0 ), the operation $ is defined as p $ q = (p-1)^(q-1). What is the value of (p +1) $ (q +1)

I. p =1

II. q = 0
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Re: For all numbers p and q (where p ≠ 0 ), the operation $ is defined  [#permalink]

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New post 10 Apr 2019, 23:38
can someone explain this ?
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For all numbers p and q (where p ≠ 0 ), the operation $ is defined  [#permalink]

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New post 11 Apr 2019, 00:01
KarthikaD wrote:
can someone explain this ?



You can redefine the question stem as p$q = (1st term - 1) raised to the power (2nd term - 1)

(p+1)$(q+1) = ((p+1) - 1)^((q+1)-1) which simplifies to p^q

So, (p+1)$(q+1) = p^q is what we need to find

Statement (1) says p=1, so the stem becomes 1^q

We know that 1 raised to any power is 1, so this is sufficient

Statement (2) says q=0, so the stem becomes p^0

We know that any number raised to the power 0 is 1, so this is sufficient

(1) and (2) are independently sufficient

(D)
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For all numbers p and q (where p ≠ 0 ), the operation $ is defined   [#permalink] 11 Apr 2019, 00:01
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For all numbers p and q (where p ≠ 0 ), the operation $ is defined

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