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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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03 Jun 2020, 01:02
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57% (02:19) correct 43% (02:15) wrong based on 41 sessions
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Competition Mode Question If \(P= \frac{A^n}{B^{2m}}\), where A is a positive integer less than or equal to 20, B is a nonzero integer between 10 and 10, inclusive, and m and n are nonnegative singledigit 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. \(10^{18} < P < 20^9\) Are You Up For the Challenge: 700 Level Questions
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Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to
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03 Jun 2020, 01:39
\(A^n\) 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
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03 Jun 2020, 03:43
If \(P=\frac{A^n}{B^{2m}}\), where A is a positive integer less than or equal to 20, B is a nonzero integer between 10 and 10, inclusive, and m and n are nonnegative singledigit 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
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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 nonzero integer between 10 and 10, inclusive, and m and n are nonnegative singledigit 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  Incorrect 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
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03 Jun 2020, 05:37
Quote: If P=AnB2m, where A is a positive integer less than or equal to 20, B is a nonzero integer between 10 and 10, inclusive, and m and n are nonnegative singledigit 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
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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
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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
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03 Jun 2020, 19:01
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, \(10^{18} < P \leq 20^{9}\)




Re: If P= A^n/B^2m, where A is a positive integer lesser than or equal to
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03 Jun 2020, 19:01




