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If P= A^n/B^2m, where A is a positive integer lesser than or equal to

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If P= A^n/B^2m, where A is a positive integer lesser than or equal to  [#permalink]

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New post 03 Jun 2020, 01:02
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If \(P= \frac{A^n}{B^{2m}}\), where A is a positive integer less than or equal to 20, B is a non-zero integer between -10 and 10, inclusive, and m and n are non-negative single-digit numbers, which of the following expressions is true?


A. \(0 < P < (0.2)^9\)

B. \(-10^{-9} < P < (0.2)^9\)

C. \(10^{-18} < P < 20^9\)

D. \(-20^{-9} <P<20^9\)

E. \(-10^{-18} < P < 20^9\)


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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to  [#permalink]

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New post 03 Jun 2020, 01:39
2
\(A^n\) and \((B^2)^m\) both are greater than 0; Hence, \(\frac{A^n}{B^{2m}}\) can never be negative.

Eliminate B, D and E

Maximum value of \(A^n/B^{2m}\) occurs, when \(A^n\) attains it's maximum value and \(B^{2m}\) attains it's minimum.

Maximum possible value of \(A^n = 20^9\)

Minimum possible value of \(B^{2m} = 1^{2*1} = 1\)

Maximum value of \(\frac{A^n}{B^{2m}} = 20^9\)

Eliminate A

C (Answer)

We don't even need to find the minimum value of P to get to the answer. Anyways,

Minimum value of \(A^n/B^{2m}\) occurs, when \(A^n\) attains it's minimum value and \(B^{2m}\) attains it's maximum.

Minimum possible value of \(A^n = 1^1\)

Maximum possible value of \(B^{2m}= 10^{2*9} = 10^{18}\)

Minimum value of \(\frac{A^n}{B^{2m}} = 1/10^{18} = 10^{-18} \)
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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to  [#permalink]

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New post 03 Jun 2020, 03:43
1
If \(P=\frac{A^n}{B^{2m}}\), where A is a positive integer less than or equal to 20, B is a non-zero integer between -10 and 10, inclusive, and m and n are non-negative single-digit numbers, which of the following expressions is true?


A. \(0<P<(0.2)^9\)

B. \(−10^{−9}<P<(0.2)^9\)

C. \(10^{−18}<P<20^9\)

D. \(−20^{−9}<P<20^9\)

E. \(−10^{−18}<P<20^9\)

\(P=\frac{A^n}{B^{2m}}\) is minimum when A and n are minimum AND B and m are maximum
If \(P=\frac{A^n}{B^{2m}}\) is maximum when A and n are maximum AND B and m are minimum.

1 ≤ A ≤ 20
-10 ≤ B ≤ -1 and 1 ≤ B ≤ 10
0 ≤ m/n ≤ 9

\(P_{min} = \frac{1^0}{10^{2*9}} = 10^{-18}\)

\(P_{max} = \frac{20^9}{1^{2*1}} = 20^9\)

\(P_{min} ≤ P ≤ P_{max} = 10^{-18} ≤ P ≤ 20^9\)

So, at least
\(10^{-18} < P < 20^9\)

Answer C.
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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to  [#permalink]

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New post 03 Jun 2020, 05:30
Is it C ?

If P=A^n/B^2m , where A is a positive integer less than or equal to 20, B is a non-zero integer between -10 and 10, inclusive, and m and n are non-negative single-digit numbers, which of the following expressions is true?

So since A >0 and (- B )^even >0 ... value of P > 0 . So BDE option out ..

Highest value of P is when denominator minimal = 1 and numerator is maximum .


A = 20 , B = 10 or -10 m and n = between 0 ..9 inclusive , P = (20^9 / 10 ^2*0 ) = 20 ^9

Lowest value of P is when denominator max = 10^2*9 and numerator is minimum = 1 ^0 . = 1
P = (1^0 / 10 ^2*9 ) = 10 ^-18


so option C matches .
A. 0<P<(0.2)^9 -- In-correct based on the condition given

B. -ve value not possible -- out

C. 10^−18<P<20^9

D. -ve value not possible -- out

E. -ve value not possible -- out
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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to  [#permalink]

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New post 03 Jun 2020, 05:37
1
Quote:
If P=AnB2m, where A is a positive integer less than or equal to 20, B is a non-zero integer between -10 and 10, inclusive, and m and n are non-negative single-digit numbers, which of the following expressions is true?

A. 0<P<(0.2)9
B. −10−9<P<(0.2)9
C. 10−18<P<209
D. −20−9<P<209
E. −10−18<P<209


P=A^n/B^2m
0<A<21 (integer)
-10<B=!0<10 (integer)
-1<m,n<10 (integer)

maximum
A=20, B=1, n=9
P=20^9/1

minimum
A=1, B=9 or -9, m=9
P=1/9^2(9)=1/9^(18)=9^(-18)

ans (C)
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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to  [#permalink]

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New post 03 Jun 2020, 07:36
E is the right answer.

Concept : Max of fraction- Nr should be max/Dr to be min

Min of fraction - Nr should be min/Dr should be max

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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to  [#permalink]

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New post 03 Jun 2020, 12:58
Is the answer to this questin C?
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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to  [#permalink]

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New post 03 Jun 2020, 19:01
1
Given: \(P=\frac{A^n}{B^{2m}}\). Since the power of B is even, the fraction will never be negative.

\(0<A\leq 20\)
\(-10\leq B\leq 10\)
\(0<m,n\leq 9\)

Min value of P will occur -
A - min = 1
B - max = -10, 10
n - min = 1
m - max = 9
\(P = 10^{-18}\)

Max value of P will occur -
A - max = 20
B - min = -1, 1
n - max = 9
m - min = 1
\(P = 20^{9}\)

So, \(10^{-18} < P \leq 20^{9}\)
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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to   [#permalink] 03 Jun 2020, 19:01

If P= A^n/B^2m, where A is a positive integer lesser than or equal to

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