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03 Oct 2022, 02:11
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
61% (02:25) correct
39%
(02:18)
wrong
based on 252
sessions
History
Date
Time
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Not Attempted Yet
If \(\frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\), then what is the value of x ?
A. 6
B. 7
C. 8
D. 12
E. 14
A. 6
B. 7
C. 8
D. 12
E. 14
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21 Feb 2023, 13:36
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Bunuel
Official Solution:
If \(\frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\), then what is the value of x ?
A. \(6\)
B. \(7\)
C. \(8\)
D. \(12\)
E. \(14\)
Get rid of the roots:
\(\frac { (100!)^{\frac{1}{2}}*(100!)^{\frac{1}{4}}*(100!)^{\frac{1}{8}} } {(99!)^{\frac{1}{2}}*(99!)^{\frac{1}{4}}*(99!)^{\frac{1}{8}} } =\sqrt[8]{10^x}\);
Take the whole expression to the eights power:
\((\frac { (100!)^{\frac{1}{2}}*(100!)^{\frac{1}{4}}*(100!)^{\frac{1}{8}} } {(99!)^{\frac{1}{2}}*(99!)^{\frac{1}{4}}*(99!)^{\frac{1}{8}} } )^8=(\sqrt[8]{10^x})^8\);
\(\frac{(100!)^4*(100!)^2*100!}{(99!)^4*(99!)^2*99!}=10^x\);
\(\frac{(100!)^7}{(99!)^7}=10^x\);
\((\frac{100!}{99!})^7=10^x\);
\(100^7=10^x\);
\(10^{14}=10^x\);
\(x=14\).
Answer: E
General Discussion
03 Oct 2022, 04:22
Kudos
Bookmarks
We can simplify the denominator first and then use that information to work the numerator.
Denominator = \(\sqrt{99!\sqrt{99!\sqrt{99!}}}\)
Assume
\(\sqrt{99!}\) = x
\(\sqrt{99! * x}\) = y
\(\sqrt{99! * y}\) = z
\(\sqrt{99!\sqrt{99!\sqrt{99!}}}\) = z
Thus the denominator can be denoted by z
Numerator= \(\sqrt{100!\sqrt{100!\sqrt{100!}}}\)
\(\sqrt{100!\sqrt{100! * \sqrt{99! * 100}}}\)
\(\sqrt{100!\sqrt{100! * \sqrt{99!}* \sqrt{100}}}\)
\(\sqrt{100!\sqrt{100! * x * 10}}\)
\(\sqrt{100!\sqrt{100 * 99! * x * 10}}\)
\(\sqrt{100!\sqrt{1000 * 99! * x}}\)
\(\sqrt{100! * 10^{\frac{3}{2}} * Y }\)
\(\sqrt{10^{7/2} * 99! * y }\)
\(10^{7/4} * z\)
LHS = \(10^{7/4} * \frac{z }{ z}\) = \(10^{7/4}\)
RHS = \(10^{x/8}\)
Comparing
\(\frac{x}{8}\) = \(\frac{7}{4}\)
x = 14
Method 2 - In fact an easier one
Raise both sides to the power of 8
\(\frac{100!^{4} * 100!^{2} * 100! }{ 99!^{4} * 99!^{2} * 99!}\) = \(10^x\)
\(\frac{100^{4} * 100^{2} * 100 * (99!^{4} * 99!^{2} * 99!) }{ 99!^{4} * 99!^{2} * 99!}\) = \(10^x\)
\(10^{14}\) = \(10^x\)
x = 14
IMO - E
Denominator = \(\sqrt{99!\sqrt{99!\sqrt{99!}}}\)
Assume
\(\sqrt{99!}\) = x
\(\sqrt{99! * x}\) = y
\(\sqrt{99! * y}\) = z
\(\sqrt{99!\sqrt{99!\sqrt{99!}}}\) = z
Thus the denominator can be denoted by z
Numerator= \(\sqrt{100!\sqrt{100!\sqrt{100!}}}\)
\(\sqrt{100!\sqrt{100! * \sqrt{99! * 100}}}\)
\(\sqrt{100!\sqrt{100! * \sqrt{99!}* \sqrt{100}}}\)
\(\sqrt{100!\sqrt{100! * x * 10}}\)
\(\sqrt{100!\sqrt{100 * 99! * x * 10}}\)
\(\sqrt{100!\sqrt{1000 * 99! * x}}\)
\(\sqrt{100! * 10^{\frac{3}{2}} * Y }\)
\(\sqrt{10^{7/2} * 99! * y }\)
\(10^{7/4} * z\)
LHS = \(10^{7/4} * \frac{z }{ z}\) = \(10^{7/4}\)
RHS = \(10^{x/8}\)
Comparing
\(\frac{x}{8}\) = \(\frac{7}{4}\)
x = 14
Method 2 - In fact an easier one
Raise both sides to the power of 8
\(\frac{100!^{4} * 100!^{2} * 100! }{ 99!^{4} * 99!^{2} * 99!}\) = \(10^x\)
\(\frac{100^{4} * 100^{2} * 100 * (99!^{4} * 99!^{2} * 99!) }{ 99!^{4} * 99!^{2} * 99!}\) = \(10^x\)
\(10^{14}\) = \(10^x\)
x = 14
IMO - E
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