fskilnik wrote:
GMATH practice question (Quant Class 16)
In the last elections, three small political parties had 90, 108 and 144 seconds for daily propaganda in television. If all three parties had different daily numbers of propaganda appearances, but all appearances took the same maximum possible number of seconds, how many daily propaganda appearances were offered in the television for these parties combined?
(A) 15
(B) 16
(C) 17
(D) 19
(E) 21
YES... this is a "GCD (=HCF) problem", no doubt! Thank you all for your nice contributions!
Here is the "official solution":
\(a,b,c\,\, \ge 1\,\,\,{\rm{ints}}\,\,\,:\,\,\,{\rm{numbers}}\,\,{\rm{of}}\,\,{\rm{daily}}\,\,{\rm{TV}}\,\,{\rm{appearances}}\)
\(t\, \ge 1\,\,{\mathop{\rm int}} \,\,\,:\,\,{\rm{max}}\,\,\left( * \right)\,\,\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{seconds/appearance}}\)
\(? = a + b + c\)
\(\left. \matrix{
a \cdot t = 90\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,2 \cdot {3^2} \cdot 5\,\,\, \hfill \cr
b \cdot t = 108\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,{2^2} \cdot {3^3} \hfill \cr
c \cdot t = 144\,\,\,\,\, \Rightarrow \,\,\,\,t\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,{2^4} \cdot {3^2} \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,t = GCD\left( {2 \cdot {3^2} \cdot 5\,\,;\,\,{2^2} \cdot {3^3}\,\,;\,\,{2^4} \cdot {3^2}} \right) = 2 \cdot {3^2}\)
\(? = {{2 \cdot {3^2} \cdot 5} \over {2 \cdot {3^2}}} + {{{2^2} \cdot {3^3}} \over {2 \cdot {3^2}}} + {{{2^4} \cdot {3^2}} \over {2 \cdot {3^2}}} = 5 + 6 + 8 = 19\)
The correct answer is therefore (D).
We follow the notations and rationale taught in the
GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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