On December 20, 2023, Brent will stop offering office hours.
- Video Course
- Video Course Overview
- General GMAT Strategies - 7 videos (free)
- Data Sufficiency - 16 videos (free)
- Arithmetic - 38 videos
- Powers and Roots - 36 videos
- Algebra and Equation Solving - 73 videos
- Word Problems - 48 videos
- Geometry - 42 videos
- Integer Properties - 38 videos
- Statistics - 20 videos
- Counting - 27 videos
- Probability - 23 videos
- Analytical Writing Assessment - 5 videos (free)
- Reading Comprehension - 10 videos (free)
- Critical Reasoning - 38 videos
- Sentence Correction - 70 videos
- Integrated Reasoning - 17 videos
- Study Guide
- Blog
- Philosophy
- Office Hours
- Extras
- Prices
Comment on Testing Possible Cases
Hi Brent,
Need your help: https://gmatclub.com/forum/if-m-p-s-and-v-are-positive-and-m-p-s-v-which-of-the-following-160298.html
I tried solving it by input-output method and I don't think that's the best approach for it. I chose the following values: m = 3; p = 5; s = 7; v = 2
Question link: https:/
Question link: https://gmatclub.com/forum/if-m-p-s-and-v-are-positive-and-m-p-s-v-which...
Yeah, the input-output method MAY help you eliminate some answer choices, but it won't take you all the way to the correct answer.
Here's my full solution: https://gmatclub.com/forum/if-m-p-s-and-v-are-positive-and-m-p-s-v-which...
Cheers,
Brent
Hey Brent :) Can you give me
My solution to question 390:
My solution to question 390: https://gmatclub.com/forum/a-school-administrator-will-assign-each-stude...
My solution to question 176: https://gmatclub.com/forum/if-n-3-8-2-8-which-of-the-following-is-not-a-...
Aside: Here are links to solutions to ALL official guide questions: https://gmatclub.com/forum/gmac-official-guides-the-master-directory-lin...
Cheers,
Brent
https://gmatclub.com/forum/if
Hi Brent. I just wanted to say thank you for all these videos and resources. If I had looked at the above linked questions before taking your course, I would have been in shock. But now I am having much much greater success.
Good to hear!!
Good to hear!!
Hi Brent,
Please review my solution.
statement 1: xy+y=ODD ; y(x+1)=Odd . This implies y= odd and x+1 = odd. If x+1=odd then x= even.(sufficient)
statement 2: 6x-3y=ODD; 3(2x-y)=odd. We know 3 is odd.(2x-y)=Odd ; x can be even or x can be odd. (Not sufficient)
So, opt A is the correct ans.
That's a perfectly-reasoned
That's a perfectly-reasoned solution. In fact, it's exactly how I'd solve such a question.
That said, if students aren't sure how to proceed with similar questions, they can always just test all possible cases as we have done above.
If r and s are integers and
A. r
B. s
C. r + s
D. rs - r
E. r^2 + s
I drew a table.there are two scenarios
When rs+r is odd...
1/r=odd,s =even,rs+r odd
2/ r=even, S= odd, rs+r odd..
Confused what to do!
That's a good approach. Let's
That's a good approach. Let's start it from the beginning by testing all four possible cases:
1) r is EVEN and s is EVEN: Here, rs + r is EVEN
2) r is EVEN and s is ODD: Here, rs + r is EVEN
3) r is ODD and s is EVEN: Here, rs + r is ODD
4) r is ODD and s is ODD: Here, rs + r is EVEN
Aha! In your solution, you say that, if r=even and s=odd, then rs+r is odd, but this is not the case.
For example, if r = 0 and s = 1, then rs + r = (0)(1) + 0 = 0, which is even.
As we can see above, case #3 is the only scenario that yields an ODD value of rs + r.
So, it must be the case that r is ODD and s is EVEN
Does that help?
https://gmatclub.com/forum/if
Basically we test four cases;but in here two cases is being tasted in your solution. Would you please explain?
Solution link: https:/
Solution link: https://gmatclub.com/forum/if-x-and-y-are-integers-is-x-even-226190.html...
Let's compare this question to the question I answered in the above comments (If r and s are integers and rs + r is odd, which of the following must be even?)
For the rs+r question, we're given the expression rs + r, and our job is to determine the circumstances in which rs + r is odd.
Notice that rs + r is an algebraic EXPRESSION (not an EQUATION).
As such, we can test all 4 possible cases (e.g., r is odd and s is even) to see which case(s) yields an odd value for rs + r
In the other question (If x and y are integers, is x even?), statement 1 tells us that x + y = y⁵
Notice that x + y = y⁵ is an EQUATION.
So, while we COULD try testing all four cases, it's more convenient to rewrite x + y = y⁵ as x = y⁵ - y, and then see what happens to x when y is ODD and when y is EVEN.
That said, we could have also tested all four cases. Let's do that.
1) x is EVEN and y is EVEN:
Here, x + y = y⁵ becomes EVEN + EVEN = EVEN⁵
Simplify: EVEN + EVEN = EVEN
This works.
2) x is EVEN and y is ODD:
Here, x + y = y⁵ becomes EVEN + ODD = ODD⁵
Simplify: EVEN + ODD = ODD
This works.
3) x is ODD and y is EVEN:
Here, x + y = y⁵ becomes ODD + EVEN = EVEN⁵
Simplify: ODD + EVEN = EVEN
Does NOT work.
4) x is ODD and y is ODD:
Here, x + y = y⁵ becomes ODD + ODD = ODD⁵
Simplify: ODD + ODD = ODD
Does NOT work.
Cases 1 and 2 are the only ones that work.
In both cases, x is EVEN, so we can answer the target question with certainty.
Does that help?
Cheers,
Brent
If n is a positive integer,
(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd
(D) divisible by 3 only when n is odd
(E) divisible by 4 whenever n is even
can you pls explain when it is divisible by 3 for e.g
when n is odd 3*4*5 = 60 its div by 3 ... then why ans E ...
Here's my full solution:
Here's my full solution: https://gmatclub.com/forum/if-n-is-a-positive-integer-then-n-n-1-n-2-is-...
Please let me know if you have any questions about my solution.
Hi Brent,
Can you please clarify on your factoring solution to this question?:
https://gmatclub.com/forum/m-and-n-are-positive-integers-if-mn-2m-n-1-is-even-265828.html
You said, "Recognize that, if mn + 2m + n + 1 is even, then mn + 2m + n + 2 must be ODD." What if we assume the other way around where mn + 2m + n + 1 is odd and mn + 2m + n + 2 would be even?
Thank you!
That's correct.
That's correct.
We know that 1 is ODD.
So, if (mn + 2m + n + 1) is EVEN, then (mn + 2m + n + 1) + 1 = EVEN + ODD = ODD
In other words, mn + 2m + n + 2 is ODD.
Conversely, if (mn + 2m + n + 1) is ODD, then (mn + 2m + n + 1) + 1 = ODD + ODD = EVEN
In other words, mn + 2m + n + 2 is EVEN.
Got it, so however we assign
That's correct.
That's correct.