How many positive two-digit numbers are odd, not divisible b

[fskilnik@GMATH]

GMATH practice exercise (Quant Class 16)

How many positive two-digit numbers are odd, not divisible by 3, and have distinct digits?

(A) 28
(B) 27
(C) 26
(D) 25
(E) 24

Answer: [spoiler]____(B)__[/spoiler]

[Brent@GMATPrepNow]

GMATH practice exercise (Quant Class 16)

How many positive two-digit numbers are odd, not divisible by 3, and have distinct digits?

(A) 28
(B) 27
(C) 26
(D) 25
(E) 24

Answer: [spoiler]____(B)__[/spoiler]

After about 30 seconds, I recognized that I couldn't see a nice straightforward solution (one that doesn't involve considering multiple cases), so I had 3 options:
1) Keep searching for a nice straightforward solution
2) Guess and move on
3) List the possible outcomes

Since the answer choices are reasonably small, I can quickly list the outcomes and be finished in under 2 minutes (total)
So, let's do that.

We get: 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 79, 83, 85, 89, 91, 95, 97

Answer: B

Cheers,
Brent

[GMATGuruNY]

GMATH practice exercise (Quant Class 16)

How many positive two-digit numbers are odd, not divisible by 3, and have distinct digits?

(A) 28
(B) 27
(C) 26
(D) 25
(E) 24

For any set of consecutive integers:
count = biggest - smallest + 1
Thus:
Number of two-digit integers between 10 and 99, inclusive = 99-10+1 = 90.

Of these 90 consecutive integers, 1 of every 3 will be a multiple of 3, implying that 2/3 will NOT be divisible by 3:
(2/3)(90) = 60.

Of these 60 remaining integers, exactly 1/2 will be ODD:
(1/2)(60) = 30.

Of these 30 remaining integers that are not divisible by 3 but are odd, three -- 11, 55 and 77 -- have repeated digits and thus must be subtracted from the total:
30-3 = 27.

The correct answer is B.

[fskilnik@GMATH]

GMATH practice exercise (Quant Class 16)

How many positive two-digit numbers are odd, not divisible by 3, and have distinct digits?

(A) 28
(B) 27
(C) 26
(D) 25
(E) 24

$$?\,\,\,\,:\,\,\,\# N\,,\,\,N \in \left[ {10,99} \right]\,\,,\,\,{\rm{odd}}\,\,{\rm{,}}\,\,{\rm{not}}\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,{\rm{3}}\,,\,\,\,{\rm{not}}\,\,{\rm{divisible}}\,\,{\rm{by}}\,\,11$$
$${\rm{I}}{\rm{.}}\,\,\,\,{\rm{odd}} \in \left[ {10,99} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{1 \over 2}\left( {99 - 10 + 1} \right) = 45\,\,{\rm{numbers}}$$

$${\rm{I}}{\rm{.}}\,\, \cap \,\,\left( {{\rm{div}}\,\,{\rm{by}}\,\,3} \right)\,\,\,:\,\,\,\left\{ \matrix{
\,15 = 3 \cdot 5 + 0 \cdot 6 \hfill \cr
\,21 = 3 \cdot 5 + 1 \cdot 6 \hfill \cr
\,\,\, \ldots \hfill \cr
\,99 = 3 \cdot 5 + 14 \cdot 6 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,14 + 1 = 15\,\,{\rm{numbers}}$$
$${\rm{I}}{\rm{.}}\,\, \cap \,\,\left( {\underline {{\rm{not}}} \,\,{\rm{div}}\,\,{\rm{by}}\,\,3} \right)\,\,\,:\,\,\,45 - 15 = 30\,\,{\rm{numbers}}$$

$${\rm{I}}{\rm{.}}\,\, \cap \,\,\left( {\underline {{\rm{not}}} \,\,{\rm{div}}\,\,{\rm{by}}\,\,3} \right) \cap \,\,\left( {{\rm{div}}\,\,{\rm{by}}\,\,11} \right)\,\,\,:\,\,\,\left\{ {11,33,55,77,99} \right\} - \left\{ {33,99} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,3\,\,{\rm{numbers}}$$

$$?\,\, = \,\,{\rm{I}}{\rm{.}}\,\, \cap \,\,\left( {\underline {{\rm{not}}} \,\,{\rm{div}}\,\,{\rm{by}}\,\,3} \right) \cap \,\,\left( {\underline {{\rm{not}}} \,\,{\rm{div}}\,\,{\rm{by}}\,\,11} \right) = \,\,30 - 3 = 27\,\,{\rm{numbers}}$$


The correct answer is (B).


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

Regards,
Fabio.