If q is a six-digit integer between 200,000 and 201,000, what is the

I'll go with option C.

(1) Units digit can be 4 or 9. Hence not sufficient.

(2) Tens digit of 10q+3 is 9. So 10th digit of 10q can be 8 or 9, likewise the unit's digit of q. Hence not sufficient.

Example XXXX89 + 3 = XXXXX92.

XXXXX90 + 3 = XXXXX93.

Combining (1) and (2) - Unit's digit is 9

Cheers!

Okay I realized I made a blunder.

(2) Tens digit of 10q+3 is 9. So 10th digit of 10q can be 8 or 9. - The last digit of 10q will always be 0. So only 9 is possible. Hence B alone is sufficient.

Cheers!

\(200,000 < q\,\,{\mathop{\rm int}} \,\, < \,\,201,000\)

\(? = \left\langle q \right\rangle\)

\(\left( 1 \right)\,\,q = 5M + 4\,\,,\,\,\,M\,\,{\mathop{\rm int}} \,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,M = 40,000\,\,\,\, \Rightarrow \,\,\,q = 200,004\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 4 \hfill \cr \\
\,{\rm{Take}}\,\,M = 40,001\,\,\,\, \Rightarrow \,\,\,q = 200,009\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 9 \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\left\langle {{{10q + 3} \over {10}}} \right\rangle \,\, = \,\,9\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {q + {3 \over {10}}} \right\rangle = 9\,\,\,\,\,\mathop \Rightarrow \limits^{q\,\,{\mathop{\rm int}} } \,\,\,\,\,? = 9\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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