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655-705 Level|   Algebra|   Exponents|   Roots|      
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brainwired
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NEW!!! Tough and tricky exponents and roots questions 23 Jul 2023, 23:31
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Bunuel said any positive int root from a number more than one is more than one.

\(\sqrt[1000]{2}\)>1 but that means it is 2^(1/1000) right? How is that greater than 1?
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Bunuel said any positive int root from a number more than one is more than one.

\(\sqrt[1000]{2}\)>1 but that means it is 2^(1/1000) right? How is that greater than 1?
Show more

\(\sqrt[1000]{2} \approx 1.000693...\)

Also, consider this: let \(\sqrt[1000]{2} = x\), where x > 0. Then \(2 = x^{1000}\). Now, if x were less than 1, then \(x^{1000}\) would be less than x (when taking a number between 0 and 1 to a positive integer power, its value decreases. For instance, (1/2)^2 = 1/4, which is less than 1/2). Hence, x must be greater than 1.

Hope it's clear.
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In question number 1, once I reach: 50 + 2 * (sqr root of (25^2 - 10 * sqr root of 6))^2 could I operate in the following way?
Pass the 25 to the left side and multiply by 2 so: 50 + 50 * (sqr root of -600)?
Then do decompose 600 and take the square out of the square root so that: 100*2*5 * (sqr root of -3)
It does not work as I do not get to the right answer... Can anyone please help me understand what I am doing wrong here?
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ruis
In question number 1, once I reach: 50 + 2 * (sqr root of (25^2 - 10 * sqr root of 6))^2 could I operate in the following way?
Pass the 25 to the left side and multiply by 2 so: 50 + 50 * (sqr root of -600)?
Then do decompose 600 and take the square out of the square root so that: 100*2*5 * (sqr root of -3)
It does not work as I do not get to the right answer... Can anyone please help me understand what I am doing wrong here?
Show more

When you factor out a term from an expression, ensure to factor that term out from every part of the expression. For example, in ax + bx, factoring out x means taking x out of both ax and bx, which gives x(a + b). Thus, you cannot just factor out 25^2 from only the first term in \(\sqrt{25^2-(10\sqrt{6})^2)}\).
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6. If \(x=\sqrt[5]{-37}\) then which of the following must be true?
A. \(\sqrt{-x}>2\)
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32

MUST KNOW FOR THE GMAT:

• Even roots from a negative number are undefined on the GMAT (as GMAT is dealing only with Real Numbers): \(\sqrt[{even}]{negative}=undefined\), for example, \(\sqrt{-25}=undefined\).

• Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

BACK TO THE ORIGINAL QUESTION:

As \(-2^5=-32\), then \(x\) must be a little bit less than -2, hence \(x=\sqrt[5]{-37} \approx -2.1 \lt -2\). Thus \(x^3 \approx (-2.1)^3 \approx -8.something \lt -8\), so option D must be true.

As for the other options:

A. \(\sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1} \lt 2\), \(\sqrt{-x} \gt 2\) is not true.

B. \(x \approx -2.1 \lt -2\), thus \(x \gt -2\) is also not true.

C. \(x^2 \approx (-2.1)^2=4.something \gt 4\), thus \(x^2 \lt 4\) is also not true.

Answer: D.
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How about answer choice E? E looks true as well as D
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Bunuel
7. If \(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}\), then which of the following must be true?

A. \(x \gt 12\)
B. \(10 \lt x \lt 12\)
C. \(8 \lt x \lt 10\)
D. \(6 \lt x \lt 8\)
E. \(x \lt 6\)


Here is a useful little trick: any positive integer root of a number greater than 1 will be more than 1. For example: \(\sqrt[1000]{2} \gt 1\).

Now, \(\sqrt{10} \gt 3\) (as \(3^2=9\)) and \(\sqrt[3]{9} \gt 2\) (as \(2^3=8\)).

Therefore:

    \(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}=\)

    \(=(\text{number more than 3})+(\text{number more than 2})+(\text{7 numbers more than 1})=\)

    \(=(\text{number more than 5})+(\text{number more than 7})=\)

    \(=(\text{number more than 12})\)


Answer: A
Answer: E.
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Is there another way to do this exercise? I tried calculating the maximun and the minimum to understand the range of x. We have the sum of 9 terms. If all terms were equal to the largest term (sqrt{10}), we would have: sum = 9* (sqrt{10}) with is aprox 27 (a bit more than 27). So, the actual sum must be less than 27.
Then, we do the same with the smallest: sum = 9* 2^(1/10).... 3^2*2^(1/10). I struggle solving that last expression...
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Bunuel
6. If \(x=\sqrt[5]{-37}\) then which of the following must be true?
A. \(\sqrt{-x}>2\)
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32

MUST KNOW FOR THE GMAT:

• Even roots from a negative number are undefined on the GMAT (as GMAT is dealing only with Real Numbers): \(\sqrt[{even}]{negative}=undefined\), for example, \(\sqrt{-25}=undefined\).

• Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

BACK TO THE ORIGINAL QUESTION:

As \(-2^5=-32\), then \(x\) must be a little bit less than -2, hence \(x=\sqrt[5]{-37} \approx -2.1 \lt -2\). Thus \(x^3 \approx (-2.1)^3 \approx -8.something \lt -8\), so option D must be true.

As for the other options:

A. \(\sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1} \lt 2\), \(\sqrt{-x} \gt 2\) is not true.

B. \(x \approx -2.1 \lt -2\), thus \(x \gt -2\) is also not true.

C. \(x^2 \approx (-2.1)^2=4.something \gt 4\), thus \(x^2 \lt 4\) is also not true.

Answer: D.


How about answer choice E? E looks true as well as D
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\(x=\sqrt[5]{-37} \approx -2.1 \lt -2\).

2^4 = 16, so (-2.1)^4 will be a little greater than 15 bur for sure less than 32.
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Bunuel
7. If \(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}\), then which of the following must be true?

A. \(x \gt 12\)
B. \(10 \lt x \lt 12\)
C. \(8 \lt x \lt 10\)
D. \(6 \lt x \lt 8\)
E. \(x \lt 6\)


Here is a useful little trick: any positive integer root of a number greater than 1 will be more than 1. For example: \(\sqrt[1000]{2} \gt 1\).

Now, \(\sqrt{10} \gt 3\) (as \(3^2=9\)) and \(\sqrt[3]{9} \gt 2\) (as \(2^3=8\)).

Therefore:

    \(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}=\)

    \(=(\text{number more than 3})+(\text{number more than 2})+(\text{7 numbers more than 1})=\)

    \(=(\text{number more than 5})+(\text{number more than 7})=\)

    \(=(\text{number more than 12})\)


Answer: A
Answer: E.

Is there another way to do this exercise? I tried calculating the maximun and the minimum to understand the range of x. We have the sum of 9 terms. If all terms were equal to the largest term (sqrt{10}), we would have: sum = 9* (sqrt{10}) with is aprox 27 (a bit more than 27). So, the actual sum must be less than 27.
Then, we do the same with the smallest: sum = 9* 2^(1/10).... 3^2*2^(1/10). I struggle solving that last expression...
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The approximation is pretty much the only way. However, you can check slight variations here: https://gmatclub.com/forum/m09-183829.html
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if sqrt (z-3y)=10, then shouldnt it be sqrt (45x10/10) and the answer as sqrt(10)
Bunuel
10. Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined

By rearranging both expressions, we get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Let's denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\), then \((a-b)(a+b)=5*(a+b)=125\), which implies that \(a+b=25\).

This gives us two equations with two unknowns: \(a+b=25\) and \(a-b=5\). Solving for \(a\) yields \(a=15=\sqrt{5x}\), so \(x=45\). Solving for \(b\) yields \(b=10=\sqrt{z-3y}\).

Finally, \(\sqrt{\frac{45(z-3y)}{x} }=\sqrt{\frac{45*10^2}{45} }=10\).


Answer: B
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if sqrt (z-3y)=10, then shouldnt it be sqrt (45x10/10) and the answer as sqrt(10)
Bunuel
10. Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined

By rearranging both expressions, we get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Let's denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\), then \((a-b)(a+b)=5*(a+b)=125\), which implies that \(a+b=25\).

This gives us two equations with two unknowns: \(a+b=25\) and \(a-b=5\). Solving for \(a\) yields \(a=15=\sqrt{5x}\), so \(x=45\). Solving for \(b\) yields \(b=10=\sqrt{z-3y}\).

Finally, \(\sqrt{\frac{45(z-3y)}{x} }=\sqrt{\frac{45*10^2}{45} }=10\).


Answer: B
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No. \(\sqrt{z-3y}=10\). So, \(z-3y=10^2\). Now, substitute \(z-3y=10^2\) into \(\sqrt{\frac{45(z-3y)}{x} } \) to get \( \sqrt{\frac{45*10^2}{45} }=10\).
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Isn’t 1973^3^2=1973^6?

Bunuel
2. What is the units digit of \((17^3)^4-1973^{3^2}\)?

A. 0
B. 2
C. 4
D. 6
E. 8


MUST KNOW FOR THE GMAT:

I. The units digit of \((abc)^n\) is the same as that of \(c^n\).


Hence, the units digit of \((17^3)^4\) is the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the same as that of \(3^{3^2}\).

II. \({(a^m)}^n=a^{mn}\) and \(a^{m^n}=a^{(m^n)}\).


Hence, \({(7^3)}^4=7^{(3*4)}=7^{12}\) and \(3^{3^2}=3^{(3^2)}=3^9\).

III. The units digit of integers in positive integer powers repeats in a specific pattern called cyclicity. The units digit of 7 and 3 in positive integer powers repeats in patterns of 4:


1. \(7^1=7\)(last digit is 7)

2. \(7^2=49\)(last digit is 9)

3. \(7^3=xx3\)(last digit is 3)

4. \(7^4=xxx1\) (last digit is 1)

5. \(7^5=xxxxxx7\) (last digit is 7 again!)

...


1. \(3^1=3\) (last digit is 3)

2. \(3^2=9\) (last digit is 9)

3. \(3^3=27\) (last digit is 7)

4. \(3^4=81\) (last digit is 1)

5. \(3^5=243\) (last digit is 3 again!)

...

Hence, the units digit of \(17^{12}\) is 1, as the 12th number in the repeating pattern 7-9-3-1 is 1, and the units digit of \(1973^9\) is 3, as the 9th number in the repeating pattern 3-9-7-1 is 3.

BACK TO THE QUESTION:

Thus, we know that the units digit of \({(17^3)}^4=17^{12}\) is 1, and the units digit of \(1973^{3^2}=1973^9\) is 3. If the first number is greater than the second number, the units digit of their difference will be 8, for example, 11 - 3 = 8. However, if the second number is greater than the first number, the units digit of their difference will be 2, for example, 11 - 13 = -2.

Consider this, even if the first number were \(100^{12}\) instead of \(17^{12}\), and the second number were \(1000^9\) instead of \(1973^9\), the first number, \(100^{12}=10^{24}\), would still be smaller than the second number \(1,000^9=10^{27}\). Therefore, \(17^{12} < 1973^9\), and the units digit of the difference is 2.

Answer B.
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Isn’t 1973^3^2=1973^6?

Bunuel
2. What is the units digit of \((17^3)^4-1973^{3^2}\)?

A. 0
B. 2
C. 4
D. 6
E. 8


MUST KNOW FOR THE GMAT:

I. The units digit of \((abc)^n\) is the same as that of \(c^n\).


Hence, the units digit of \((17^3)^4\) is the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the same as that of \(3^{3^2}\).

II. \({(a^m)}^n=a^{mn}\) and \(a^{m^n}=a^{(m^n)}\).


Hence, \({(7^3)}^4=7^{(3*4)}=7^{12}\) and \(3^{3^2}=3^{(3^2)}=3^9\).

III. The units digit of integers in positive integer powers repeats in a specific pattern called cyclicity. The units digit of 7 and 3 in positive integer powers repeats in patterns of 4:


1. \(7^1=7\)(last digit is 7)

2. \(7^2=49\)(last digit is 9)

3. \(7^3=xx3\)(last digit is 3)

4. \(7^4=xxx1\) (last digit is 1)

5. \(7^5=xxxxxx7\) (last digit is 7 again!)

...


1. \(3^1=3\) (last digit is 3)

2. \(3^2=9\) (last digit is 9)

3. \(3^3=27\) (last digit is 7)

4. \(3^4=81\) (last digit is 1)

5. \(3^5=243\) (last digit is 3 again!)

...

Hence, the units digit of \(17^{12}\) is 1, as the 12th number in the repeating pattern 7-9-3-1 is 1, and the units digit of \(1973^9\) is 3, as the 9th number in the repeating pattern 3-9-7-1 is 3.

BACK TO THE QUESTION:

Thus, we know that the units digit of \({(17^3)}^4=17^{12}\) is 1, and the units digit of \(1973^{3^2}=1973^9\) is 3. If the first number is greater than the second number, the units digit of their difference will be 8, for example, 11 - 3 = 8. However, if the second number is greater than the first number, the units digit of their difference will be 2, for example, 11 - 13 = -2.

Consider this, even if the first number were \(100^{12}\) instead of \(17^{12}\), and the second number were \(1000^9\) instead of \(1973^9\), the first number, \(100^{12}=10^{24}\), would still be smaller than the second number \(1,000^9=10^{27}\). Therefore, \(17^{12} < 1973^9\), and the units digit of the difference is 2.

Answer B.
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No. Please check the highlighted part.

1973^3^2 = 1973^(3^2) = 1973^9.

If it were (1973^3)^2, then it would equal (1973^3)^2 = 1973^(3 * 2) = 1973^6.
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HOW DID WE GET 25^2 AS 2 ^ ROOT Y????

Bunuel
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\), what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}\)?

A. \(\frac{14}{5}\)
B. 5
C. \(\frac{28}{5}\)
D. 13
E. 14


The first thing one should notice here is that \(x\) and \(y\) must be some irrational numbers. This is because 4,900 has primes other than 5 in its prime factorization, and 25 doesn't have 2 as a prime at all. Thus, we should manipulate the given expressions rather than solve for \(x\) and \(y\) directly.

\(5^{10x}=4,900\)

\((5^{5x})^2=70^2\)

\(5^{5x}=70\)

Now, \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}} = 5^{5x}*5^{-5}*(2^{\sqrt{y}})^2\)

Since \(5^{5x}=70\), then \(5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.
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HOW DID WE GET 25^2 AS 2 ^ ROOT Y????

Bunuel
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\), what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}\)?

A. \(\frac{14}{5}\)
B. 5
C. \(\frac{28}{5}\)
D. 13
E. 14


The first thing one should notice here is that \(x\) and \(y\) must be some irrational numbers. This is because 4,900 has primes other than 5 in its prime factorization, and 25 doesn't have 2 as a prime at all. Thus, we should manipulate the given expressions rather than solve for \(x\) and \(y\) directly.

\(5^{10x}=4,900\)

\((5^{5x})^2=70^2\)

\(5^{5x}=70\)

Now, \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}} = 5^{5x}*5^{-5}*(2^{\sqrt{y}})^2\)

Since \(5^{5x}=70\), then \(5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.
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We are given that \(2^{\sqrt{y}}=25\) and we also got that \(5^{5x}=70\), so:

\(5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2\)
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NEW!!! Tough and tricky exponents and roots questions 06 Nov 2025, 00:26
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where is the answer to question 9?
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NEW!!! Tough and tricky exponents and roots questions 06 Nov 2025, 00:35
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where is the answer to question 9?
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