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

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

by fskilnik@GMATH » Mon Feb 18, 2019 8:13 am
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

Answer: [spoiler]_____(D)__[/spoiler]
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by GMATGuruNY » Mon Feb 18, 2019 5:15 pm
fskilnik@GMATH wrote: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
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.

The correct answer is D.
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If M, N are integers greater than 1 such that 2M<N, which

by fskilnik@GMATH » Tue Feb 19, 2019 6:33 am
fskilnik@GMATH wrote: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
Although the problem seems hard, the focused-numbers (half the numbers given) are SMALL... and that´s the hint to try some "organized manual work"!

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://www.beatthegmat.com/a-number-is ... tml#827086
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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