If z is a three-digit positive integer, what is the value of

\(\underline a \,\,\underline b \,\,\underline c \,\,\,\left\{ \matrix{\\
\,\,a\,\, \in \,\,\,\left\{ {\,1,2, \ldots ,9\,} \right\} \hfill \cr \\
\,\,b\,\, \in \,\,\,\left\{ {\,0,1,2, \ldots ,9\,} \right\} \hfill \cr \\
\,\,c\,\, \in \,\,\,\left\{ {\,0,1,2, \ldots ,9\,} \right\} \hfill \cr} \right.\)

\(? = b\)


\(\left( 1 \right)\,\,\, \Rightarrow \,\,\,\left\{ \matrix{\\
\,\underline a \,21 - 91 = {\rm{tens}}\,\,{\rm{digit}}\,\,3\,\,\,\,;\,\,\, \ldots \,\,\,;\,\,\,\,\,\underline a \,29 - 91 = {\rm{tens}}\,\,{\rm{digit}}\,\,3\,\,\,\,;\,\,\,\,\,\underline a \,30 - 91 = \,\,{\rm{tens}}\,\,{\rm{digit}}\,\,3 \hfill \cr \\
{\rm{where}}\,\,a\,\, \in \,\,\,\left\{ {\,1,2, \ldots ,9\,} \right\}\,\,\, \hfill \cr} \right.\,\,\,\)

\(\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,2,1} \right)\,\,\,\, \Rightarrow \,\,\,\,\,? = \,\,2\,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,3,0} \right)\,\,\,\, \Rightarrow \,\,\,\,\,? = \,\,3\,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,\, \Rightarrow \,\,\,c = 6\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,1,6} \right)\,\,\,\, \Rightarrow \,\,\,\,\,? = \,\,1\,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,2,6} \right)\,\,\,\, \Rightarrow \,\,\,\,\,? = \,\,2\,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,c = 6\,\,\,\,\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\underline a \,26 - 91 = {\rm{tens}}\,\,{\rm{digit}}\,\,3\,\,\,\,\left( {a\,\, \in \,\,\,\left\{ {\,1,2, \ldots ,9\,} \right\}\,} \right)\,\,\,\,\, \Rightarrow \,\,\,\,b = 2\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Have to think very fast in such questions

from 1) 221 - 91, or 230 - 91

need to look for a Tens digit value as 3, we are getting 2 values here.

from 2) this is clearly insufficient, z +9 = 146 + 9, 176 +9, 186+9

Combination is a trick, But since i know from 2, UD of z has to end on a 6

for value of the tens digit of z, We can always get the value as 2

C

From 1) if the unit digit of z is >=1, the tens digit is 2 (121-91=30) and if the unit digit of z is 0, tens digit is 3 (130-91=39). Insufficient

From 2) z+9 unit digit is 5 only gives that unit digit of z is 6 (6+9=15, 16+9=25, so on). Insufficient.

From 1+2) we know unit digit is 6, i.e. >=1, therefore tens digit of z is 2. Sufficient.

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