How many positive five-digit integers contain the digit grouping "57" (in that order) at least once? For instance, 30,457 and 20,574 are two such integers to include, but 30,475 and 20,754 do not meet the restrictions.
A. 279
B. 3,000
C. 3,500
D. 3,700
E. 4,000
The OA is D.
Source: Manhattan Prep
How many positive five-digit integers contain the digit
This topic has expert replies
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
I believe that the prompt above is outdated; regardless, it does not list a correct answer. The prompt below appears in Manhattan's 5 lb. Book of GRE Practice Problems:
Number of options for the hundreds digit = 10. (Any digit 0-9.)
Number of options for the tens digit = 10. (Any digit 0-9.)
Number of options for the units digit = 10. (Any digit 0-9.)
To combine these options, we multiply:
10*10*10 = 1000.
Remaining cases: X57XX, XX57X, XXX57
Number of possible positions for the "57" grouping = 3. (The 3 cases above.)
Number of options for the leftmost digit = 9. (Any digit but 0.)
Number of options for the rightmost digit = 10. (Any digit 0-9.)
Number of options for the last remaining digit = 10. (Any digit 0-9.)
To combine these options, we multiply:
3*9*10*10 = 2700.
Total ways = Case 1 + Case 2 = 1000 + 2700 = 3700.
The correct answer is D.
Case 1: 57XXXHow many times does the digit grouping "57" (in that order) appear in all of the five-digit positive integers? For instance, "57" appears once in 12,357, twice in 57,057, and does not appear in 24,675.
A. 279
B. 3,000
C. 3,500
D. 3,700
E. 4,000
Number of options for the hundreds digit = 10. (Any digit 0-9.)
Number of options for the tens digit = 10. (Any digit 0-9.)
Number of options for the units digit = 10. (Any digit 0-9.)
To combine these options, we multiply:
10*10*10 = 1000.
Remaining cases: X57XX, XX57X, XXX57
Number of possible positions for the "57" grouping = 3. (The 3 cases above.)
Number of options for the leftmost digit = 9. (Any digit but 0.)
Number of options for the rightmost digit = 10. (Any digit 0-9.)
Number of options for the last remaining digit = 10. (Any digit 0-9.)
To combine these options, we multiply:
3*9*10*10 = 2700.
Total ways = Case 1 + Case 2 = 1000 + 2700 = 3700.
The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
- fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
The solution posted above is related to another problem (as mentioned).swerve wrote:How many positive five-digit integers contain the digit grouping "57" (in that order) at least once? For instance, 30,457 and 20,574 are two such integers to include, but 30,475 and 20,754 do not meet the restrictions.
A. 279
B. 3,000
C. 3,500
D. 3,700
E. 4,000
Source: Manhattan Prep
Let´s solve the one originally proposed, in which there are double-counting´s to be dealt with!
$$?\,\,:\,\,5{\rm{ - digit}}\,\,{\rm{positive}}\,\,{\rm{integers}}\,\,{\rm{with}}\,\,57{\rm{ - block}}\left( {\rm{s}} \right)$$
$$\eqalign{
& \left( {\rm{1}} \right)\,\,\,\underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline {} \,\,\, \underline {} \,\,\,\,\, \to \,\,\,\,{10^3}\,\,{\rm{ways}} \,\, \cr
& \left( {\rm{2}} \right)\,\,\,\underline {{\rm{not}}\,0} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline {} \,\,\,\,\, \to \,\,\,\,9 \cdot {10^2}\,\,{\rm{ways}} \,\, \cr
& \left. \matrix{
\left( {\rm{3}} \right)\,\,\,\underline {{\rm{not}}\,0} \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\,\,\, \to \,\,\,\,9 \cdot {10^2}\,\,{\rm{ways}} \hfill \cr
\left( - \right)\,\,\,\underline {\rm{5}} \,\,\, \underline 7 \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\,\,\, \to \,\,\,\,10\,\,{\rm{ways}} \hfill \cr} \right\}\,\,\,\, \to \,\,\,\,\,890\,\,{\rm{ways}} \,\, \cr
& \left. \matrix{
\left( {\rm{4}} \right)\,\,\,\underline {{\rm{not}}\,0} \,\,\, \underline {} \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, \to \,\,\,\,9 \cdot {10^2}\,\,{\rm{ways}} \hfill \cr
\left( - \right)\,\,\,\underline {{\rm{not}}\,0} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, \to \,\,\,\,9\,\,{\rm{ways}} \hfill \cr
\left( - \right)\,\,\,\underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, \to \,\,\,\,10\,\,{\rm{ways}} \hfill \cr} \right\}\,\,\,\, \to \,\,\,\,\,881\,\,{\rm{ways}} \cr} $$
$$? = 1000 + 900 + 890 + 881 = 3671$$
The alternative choices are therefore all wrong.
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br