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GMATH practice exercise (Quant Class 17)
A number is called "mystic" if it can be expressed as the sum of at least three consecutive positive integers greater than 1, in which the larger addend is more than twice the smaller addend. Note that 25 (=3+4+5+6+7) and 44 (=2+3+4+5+6+7+8+9) are mystic numbers (7>2*3 and 9>2*2). Which of the following is true?
I. 124 is mystic
II. 125 is mystic
III. 126 is mystic
(A) I. only
(B) II. only
(C) III. only
(D) Exactly two of them
(E) None of them
P.S.: I have edited this post once: the previous word "parcel" was substituted by the proper word "addend".
A number is called "mystic" if it can be expressed as the sum of at least three consecutive positive integers greater than 1, in which the larger addend is more than twice the smaller addend. Note that 25 (=3+4+5+6+7) and 44 (=2+3+4+5+6+7+8+9) are mystic numbers (7>2*3 and 9>2*2). Which of the following is true?
I. 124 is mystic
II. 125 is mystic
III. 126 is mystic
(A) I. only
(B) II. only
(C) III. only
(D) Exactly two of them
(E) None of them
P.S.: I have edited this post once: the previous word "parcel" was substituted by the proper word "addend".
20 Feb 2019, 12:52
Kudos
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fskilnik
\(?\,\,\,:\,\,\,M + \left( {M + 1} \right) + \left( {M + 2} \right) + \ldots + N\)
\(M \ge 2\)
\(N > 2M\,\,\,\,\,\left[ { \Rightarrow \,\,\,\,\,N > 4\,\,\,\,\,\mathop \Rightarrow \limits^{N\,\,{\mathop{\rm int}} } \,\,\,\,\,N \ge 5} \right]\)
\({{N\left( {N + 1} \right)} \over 2}\,\,\,\mathop = \limits^{{\rm{arithm}}{\rm{.}}\,{\rm{seq}}{\rm{.}}} \,\,\,\underbrace {1 + 2 + \ldots + \left( {M - 1} \right)}_{{{M\left( {M - 1} \right)} \over 2}\,\,\,\,\left[ {{\rm{arith}}{\rm{.}}\,{\rm{seq}}{\rm{.}}} \right]} + M + \left( {M + 1} \right) + \ldots + N\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = {{N\left( {N + 1} \right)} \over 2} - {{M\left( {M - 1} \right)} \over 2}\)
\(2\,\, \cdot \,\,?\,\,\, = \,\,\,{N^2} + N - {M^2} + M = \left( {N - M} \right)\left( {N + M} \right) + N + M = \left( {N + M} \right)\left( {N - M + 1} \right)\)
\(\left\{ \matrix{\\
\,N - M + 1 \ge 3\,\,\,\,\left[ { \ge \,\,{\rm{3}}\,\,{\rm{parcels}}\,\,\left( {{\rm{stem}}} \right)\,\,{\rm{,}}\,\,{\rm{fingers}}\,\,{\rm{trick}}} \right] \hfill \cr \\
\,N + M > N - M + 1\,\,\,\left[ {M > 0.5} \right] \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\min \left( {N + M,N - M + 1} \right) = N - M + 1 \ge 3\,\,\,\,\left( * \right)\)
\(\left( {**} \right)\,\,\left\{ \matrix{\\
\,N + M\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,\,N - M = \underbrace {N + M}_{{\rm{even}}} - \underbrace {2M}_{{\rm{even}}}\,\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,\,N - M + 1\,\,{\rm{odd}} \hfill \cr \\
\,N + M\,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\,N - M = \underbrace {N + M}_{{\rm{odd}}} - \underbrace {2M}_{{\rm{even}}}\,\,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\,N - M + 1\,\,{\rm{even}} \hfill \cr} \right.\)
\(2\,\, \cdot \,\,?\,\,\, = \,\,\,\left( {N + M} \right)\left( {N - M + 1} \right)\,\,\,{\rm{is}}\,\,{\rm{given}}\,\,{\rm{by}}\,\,\,{\rm{even}} \cdot {\rm{odd}}\,\,\left( {**} \right),\,\,{\rm{where}}\,\,\left( * \right)\,\,\,{\rm{odd}} \ge {\rm{3}}\,{\rm{,}}\,\,{\rm{even}}\,\, \ge 4\,\,\,\,\,\left( {***} \right)\)
\({\rm{I}}{\rm{.}}\,\,\,124\,\,\,\, \Rightarrow \,\,\,\,2 \cdot 124 = 248 = {2^3} \cdot 31\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,\left\{ \matrix{\\
\,{\rm{odd}} = 31 \hfill \cr \\
\,{\rm{even}} = {2^3} \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{\\
\,N + M = 31 \hfill \cr \\
\,N - M + 1 = {2^3} \hfill \cr} \right.\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\, \ldots \,\,\, \Rightarrow \,\,\,\left( {N,M} \right) = \left( {19,12} \right)\)
\(124 = 12 + 13 + \ldots + 18 + 19\,\,\,{\rm{but}}\,\,{\rm{impossible}}\,\,\,\left( {N < 2M} \right)\)
\({\rm{II}}{\rm{.}}\,\,\,125\,\,\,\, \Rightarrow \,\,\,\,2 \cdot 125 = 250 = 2 \cdot {5^3}\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,{\rm{one}}\,\,{\rm{option}}\,\,:\,\,\left\{ \matrix{\\
\,{\rm{odd}} = {5^2} \hfill \cr \\
\,{\rm{even}} = 2 \cdot 5 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{\\
\,N + M = 25 \hfill \cr \\
\,N - M + 1 = 10 \hfill \cr} \right.\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\, \ldots \,\,\, \Rightarrow \,\,\,\left( {N,M} \right) = \left( {17,8} \right)\)
\(125 = 8 + 9 + \ldots + 16 + 17\,\,\,{\rm{is}}\,\,{\rm{viable}}!\,\,\,\left( {N > 2M} \right)\)
\({\rm{III}}{\rm{.}}\,\,\,126\,\,\,\, \Rightarrow \,\,\,\,2 \cdot 126 = 252 = {2^2} \cdot {3^2} \cdot 7\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,{\rm{one}}\,\,{\rm{option}}\,\,:\,\,\left\{ \matrix{\\
\,{\rm{odd}} = 3 \cdot 7 \hfill \cr \\
\,{\rm{even}} = {2^2} \cdot 3 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\mathop \ldots \limits^{{\rm{Do}}\,\,{\rm{it}}!} \,\,\,\, \Rightarrow \,\,\,\,\left( {N,M} \right) = \left( {16,5} \right)\)
\(126 = 5 + 6 + \ldots + 15 + 16\,\,\,{\rm{is}}\,\,{\rm{viable}}!\,\,\,\left( {N > 2M} \right)\)
YES, this is a very hard problem, but I believe many "pieces" of this solution are VERY useful for candidates aiming really outstanding performances!
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
20 Feb 2019, 13:24
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Bookmarks
Hello fakilnik, I totally agree to you but I have an easy solution for this particular problem. Trust me on this I got the same answers as yours but I did this orally.
As we can see that they have already done some work for us i.e. provided us with the sum from 2 to 9 i.e. 44.
So what I did is I kept on adding number to see if I can get near to the options.
So adding from 2 to 15 = 119
So I added from 2 to 16 = 135
Now start removing numbers from the beginning of the sequence to make nearest possible mystic number.
So I started removing 2 then 3 then 4. So the sum from 5 to 16 gives me 126, a mystic number.
Now I added 17 to the new series and started removing numbers from the start removed 5,6&7 giving me a difference of -1, so the second mystic number is 125.
We cannot proceed ahead as if we'll now remove 8 and add 18 to the list then the cond. for mystic number i.e 9*2 should be greater than max is not satisfied.
May sound like a long solution but it took me 2 mins to solve.
Posted from my mobile device
As we can see that they have already done some work for us i.e. provided us with the sum from 2 to 9 i.e. 44.
So what I did is I kept on adding number to see if I can get near to the options.
So adding from 2 to 15 = 119
So I added from 2 to 16 = 135
Now start removing numbers from the beginning of the sequence to make nearest possible mystic number.
So I started removing 2 then 3 then 4. So the sum from 5 to 16 gives me 126, a mystic number.
Now I added 17 to the new series and started removing numbers from the start removed 5,6&7 giving me a difference of -1, so the second mystic number is 125.
We cannot proceed ahead as if we'll now remove 8 and add 18 to the list then the cond. for mystic number i.e 9*2 should be greater than max is not satisfied.
May sound like a long solution but it took me 2 mins to solve.
Posted from my mobile device
