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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Squaring both sides of \(|m – n| = |m| - |n|\) yields
\(|m-n|^2=(|m|-|n|)^2\)
\(=> (m-n)^2=(|m|-|n|)^2\)
\(=> m^2+n^2-2mn=|m|^2+|n|^2-2|mn|\)
\(=> m^2+n^2-2mn=m^2+n^2-2|mn|\)
\(=> -2mn=-2|mn|\)
\(=> mn=|mn|\)
\(=> mn ≥ 0\)

Condition 1):
\(m – n = 0\) implies that \(m = n.\) So \(mn = n^2 ≥ 0\) for all values of \(n\).
Condition 1) is sufficient.

Condition 2):
If \(n = 0\), then \(0 = mn ≥ 0.\)
Condition 2) is sufficient.
Therefore, the correct answer is D.
Answer: D

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Asking for \(\frac{n(n+1)(n+2)}{4}\) to be an even integer is equivalent to asking for \(n(n+1)(n+2)\) to be a multiple of \(8\). If \(n\) is an even integer, \(n\) and \(n+2\) are consecutive even integers and a product of two consecutive even integers is a multiple of \(8\). Thus, condition 1) is sufficient.

Condition 2)
If \(n = 1238, n(n+1)(n+2)=1238*1239*1240\) is a multiple of \(8\) since \(1240\) is a multiple of \(8\).
If \(n = 1239, n(n+1)(n+2)=1239*1240*1241\) is a multiple of \(8\) since \(1240\) is a multiple of \(8\).
If \(n = 1240, n(n+1)(n+2)=1240*1241*1242\) is a multiple of \(8\) since \(1240\) is a multiple of \(8\).
Thus, condition 2) is sufficient.

Therefore, the answer is D.
Answer: D

Note: This question is a CMT4(B) question. Condition 1) is easy to understand and condition 2) is hard. When one condition is easy to understand, and the other is hard, D is most likely to be the answer.

[Math Revolution GMAT math practice question]

(number property) \(f(x)\) is the greatest prime factor of \(x\). If \(n\) is a positive integer less than \(10\), what is the value of \(n\)?

\(1) f(n) = f(1000)\)
\(2) f(n) = 5\)

You are right.
The question should be changed to the followings.

If \(n\) is a positive integer, what is the value of \(n\)?

Thank you for your comments.

[Math Revolution GMAT math practice question]

(number properties) If \(n\) is a positive integer, is \(\sqrt{n+1}\) an even integer?

1) \(n\) is the product of \(2\) consecutive odd numbers
2) \(n\) is an odd number

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have \(1\) variable (\(n\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
Since \(1000 = 2^3*5^3, f(1000) = 5.\)
So, \(f(n) = 5\) and \(n = 5.\)
Thus, condition 1) is sufficient, since it gives a unique solution.

Condition 2)
Condition 2) is the same as condition 1).
Thus, condition 2) is also sufficient.

FYI, Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.

Therefore, D is the answer.
Answer: D

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

[Math Revolution GMAT math practice question]

(number properties) If \(m\) and \(n\) are positive integers, is \(m + n\) an odd number?

1) \(\frac{m}{n}\) is an even number
2) \(m\) or \(n\) is an even number

#1:
m/n = even no
m=6, n= 3 m/n= 2 ; m+n= 6+3= 9 sufficient

m=4, n=2 ; m/n= 2 ; m+n=4+2 = 6 not sufficient

#1 not sufficient

#2 :
m or n is an even no

property of odd no in addition : even+ odd = odd no
so since m & n are + integers so either of m or n being even other has to be odd integer hence we would get m+n= odd only

sufficient

IMO B

\(m,n\,\,\, \geqslant 1\,\,\,{\text{ints}}\,\,\,\,\left( * \right)\)

\(m + n\,\,\,\,\mathop = \limits^? \,\,{\text{odd}}\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\boxed{\,\,\,?\,\,\,:\,\,\,\left( {m\,\,{\text{odd}}\,,\,\,n\,\,{\text{even}}} \right)\,\,\,{\text{or}}\,\,\,{\text{vice - versa}\,\,}\,\,}\)


\(\left( 1 \right)\,\,\,\frac{m}{n} = {\text{even}}\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {4,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,m\,\,{\text{even}}\,\,\,{\text{or}}\,\,\,n\,\,{\text{even}}\,\,\,\,\left\{ \begin{gathered}\\
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\\\
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,\left( {m,n} \right) = \left( {4,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\text{E}} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

P.S.: "A or B" means "only A", "only B" or BOTH.

Beautiful problem, Max. Congrats (and kudos)!

\(n \geqslant 1\,\,\,\operatorname{int}\)

\(\sqrt {n + 1} \,\,\,\mathop = \limits^? \,\,{\text{even}}\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\boxed{\,\,n + 1\,\,\,\mathop = \limits^? \,\,\,{{\left( {{\text{even}}} \right)}^2}\,\,}\)


\(\left( 1 \right)\,\,\,n = \left( {2M - 1} \right)\left( {2M + 1} \right) = {\left( {2M} \right)^2} - {\left( 1 \right)^2}\,\,\,\,\,\left[ {M\,\,\operatorname{int} \,} \right]\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,n + 1 = {\left( {2M} \right)^2}\,\,\,,\,\,\,\,M\,\,\operatorname{int} \,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\)

\(\left( 2 \right)\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,n = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,n = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

fskilnik

#2 says m or n even ; why have you taken both as even while proving statement as insufficient?

Hi Archit3110 ,

Thank you for your interest in my solution.

As I explained in my post scriptum (PS), the word "OR" has not the everyday common use of "exclusive or".

In other words, when it is given that "m is even or n is even", there are three possibilities available:

(i) m is even and n is not even
(ii) m is not even and n is even
(iii) m is even and n is also even

Regards and success in your studies,
Fabio.

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question is equivalent to asking if \(\sqrt{n+1} = 2k\) for some positive integer \(k\).
\(\sqrt{n+1} = 2k\)
\(=> n+1 = 4k^2\)
\(=> n = 4k^2-1\)
\(=> n = (2k-1)(2k+1)\)
\(n\) is a product of two consecutive odd integers.
Thus, condition 1) is sufficient.

Condition 2)
If \(n = 3\), then \(\sqrt{3+1} = \sqrt{4}=2\) and the answer is ‘yes’.
If \(n = 1\), then \(\sqrt{1+1} = 2\) is not an integer and the answer is ‘no’.
Condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

[Math Revolution GMAT math practice question]

(number properties) If \(m\) and \(n\) are positive integers, is \(3^{4m+2}+n\) divisible by \(5\)?

\(1) m=3\)
\(2) n=1\)

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have \(2\) variables (\(m\) and \(n\)) and \(0\) equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since \(\frac{m}{n}\) is even, \(\frac{m}{n} = 2k\) and \(m = (2k)n\) for some positive integer \(k\).
If \(m = 2\) and \(n = 1\), then \(m + n = 3\), is an odd integer and the answer is ‘yes’.
If \(m = 4\) and \(n = 2\), then \(m + n = 6\) is an even integer and the answer is ‘no’.
Since we do not obtain a unique answer, conditions 1) & 2) are not sufficient when considered together.

Therefore, E is the answer.
Answer: E

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

[Math Revolution GMAT math practice question]

(function) In the \(xy\)-coordinate plane, does \(y=a(x-h)^2+k\) intersect the \(x\)-axis?

\(1) h=1\)
\(2) k=2\)

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The units digits of \(3^k\) have period \(4\) as they form the cycle \(3 -> 9 -> 7 -> 1.\)
\(3^{4m+2}\) has \(9\) as its units digit if \(3^{4m+2}\) has units digit \(9\), regardless of the value of \(m\).
Thus, the divisibility of \(3^{4m+2}+n\) by \(5\) relies on the variable n only.

Therefore, the correct answer is B.
Answer: B

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have \(3\) variables (\(a, h\) and \(k\)) and \(0\) equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
If \(a = 1\), then the graph doesn’t intersect the \(x\)-axis shown as below.



If \(a = -1\), then the graph intersects the \(x\)-axis shown as below.



Since neither condition gives us information about the value of \(a\), conditions 1) & 2) are not sufficient, when considered together.

Therefore, the answer is E.
Answer: E

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

[Math Revolution GMAT math practice question]

(absolute value) Is \(\sqrt{(x+1)^2}=x+1\) ?

\(1) x(x-2) = 0\)
\(2) x(x+2) = 0\)