If M, N are integers greater than 1 such that 2M

Question

GMATH  practice exercise (Quant Class 17)

If M, N are integers greater than 1 such that 2M<N, which of the following numbers could be  twice  the value of the sum of all integers from M to N, including both of them?

I. 54
 II. 52
III. 50

(A) I. only
(B) II. only
(C) III. only
(D) Exactly two of them
(E) None of them

The tricky alternative choice would be "All of them". We believe the examiner would NOT put it among the answer choices!

KanishkM · Answer

Would still keep this, though brute force is a time consuming approach.

2M<N, N >2M, one thing to observe was that, the series will always start from an even Integer

Will start from a single digit, 8,9,10,11,12 => 50

Now we can think greater values now

12,13,14,15 => 54

D

GMATGuruNY · Answer

54, 52 and 50 are options for TWICE the sum.
Thus, I, II, and III imply the following options for the ACTUAL SUM:
I: 27
II: 26
III: 25

Case 1: M=2
Since it is required that N>2M, N≥5
M=2 and N=5 --> sum = 2+3+4+5 = 14
M=2 and N=6 --> sum increases by 6 --> 14+6 = 20
 M=2 and N=7 --> sum increases by 7 --> 20+7 = 27 
The green case indicates that M=2 and N=7 will yield a sum of 27, implying that option I -- 54 -- can be the value of TWICE the sum.

Case 2: M=3
Since it is required that N>2M, N≥7
 M=3 and N=7 --> sum = 3+4+5+6+7 = 25 
The green case indicates that M=3 and N=7 will yield a sum of 25, implying that option III -- 50 -- can be the value of TWICE the sum.

Thus, I and III are possible.

D

OhsostudiousMJ · Answer

Hi, this is my first answer. Kindly check and suggest if my approach is right.

Given, 2M<N => Assuming the AP has only 2 numbers in the series, M and N, M+N>3M.
Thus solving 3M with the options (54/2, 52/2, 50/2) should give a reminder, suggesting that the sum of M+N is greater than the values.

That is, 
3M=27, No remainder. Therefore, M+N not greater than 3M.
3M=26, Remainder = 2. Therefore, M+N>3M.
3M=25, Remainder = 1. Therefore, M+N>3M.

Hence, (ii) and (iii) could be the answers, hence  D .

fskilnik · Answer

Hi  OhsostudiousMJ  ,

Welcome to our GMAT Club community!

You have discussed whether M+N is greater than 3M, but this is not our focus. This problem is hard, even in terms of understanding what is going on...

Suggestion: start dealing with easier questions (600-700 or below 600). You can find each question´s level looking at the questions tags when they are posted.

Regards and success in your studies!
Fabio.

fskilnik · Answer

Although the problem seems hard, the focused-numbers (half the numbers given) are SMALL... and that´s the hint to try the "organized manual work" (chosen in previous solutions)! We are looking for the 25 (III), 26 (II) and 27 (I) possibilities. {S_K} = 1 + 2 + \ldots + K = {{K\left( {K + 1} \right)} \over 2}\,\,\,\,\,\,\,\,\left( {{\rm{arithmetic}}\,\,{\rm{sequence}}} \right) {S_7} = 7 \cdot 4 = 28\,\,\, \Rightarrow \,\,\,\left\{ \matrix{ \,27 = 28 - 1 = \left( {1 + 2 + \ldots + 7} \right) - 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {N,M} \right) = \left( {7,2} \right)\,\,\,{\rm{with}}\,\,N > 2M\,\,\, \Rightarrow \,\,\,\,{\rm{viable!}} \hfill \cr \,25 = 28 - 3 = \left( {1 + 2 + \ldots + 7} \right) - \left( {1 + 2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {N,M} \right) = \left( {7,3} \right)\,\,\,{\rm{with}}\,\,N > 2M\,\,\, \Rightarrow \,\,\,\,{\rm{viable!}} \hfill \cr} \right. From the alternative choices given, we are sure we have found (EASILY!) the right answer! The correct answer is (D). We follow the notations and rationale taught in the GMATH method. Regards, Fabio. POST-MORTEM: Question 1 .: how can we be sure that 26 (II) cannot be obtained? Please note that: \left. \matrix{ {S_8} = 4 \cdot 9 = 36 \hfill \cr {S_4} = 2 \cdot 10 = 10\,\,\, \hfill \cr} \right\}\,\,\, \Rightarrow \,\,\,26 = 36 - 10 = \left( {1 + 2 + \ldots + 7 + 8} \right) - \left( {1 + 2 + 3 + 4} \right) and the ONLY reason (N,M) = (8,5) must be refuted is the fact that N>2M is false... What about other values for (N,M)? How can we be SURE there is not a single viable possibility? Question 2 .: which mathematical (GMAT-focused) properties could be useful to find all possible (N,M) pairs for a given LARGER value? We will address both questions in our very next problem, here: https://gmatclub.com/forum/a-number-is- ... l#p2229235

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If M, N are integers greater than 1 such that 2M<N, which of the foll

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions