Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 21 Oct 2013
Posts: 433

If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
29 Jun 2014, 02:44
Question Stats:
75% (02:04) correct 25% (01:57) wrong based on 283 sessions
HideShow timer Statistics
If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer? A. 5 B. 7 C. 9 D. 11 E. 13 OE M = 60·61·62·63·64·65·66·67·68·69·70 > 60 = 2·2·3·5 > 62 = 2·31 > 63 = 3·3·7 > 64 = 2·2·17 > 66 = 2·3·11 > 68 = 2·2·17 > 69 = 3·23 > 70 = 2·35
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 47946

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
29 Jun 2014, 04:54




Intern
Joined: 22 Jun 2013
Posts: 37

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
30 Jun 2014, 02:25
Hello Bunuel I tried a different approach. Is it correct ? Since we have to find the number of times 6 occurs between the products of 71 & 59. I Found the number of 3s that come till 70! Minus the number of 3s that come uptill 59!. No of 3s in 70! = 32 & No of 3s till 59! = 27 Subtracting them I got 5, which is the OA. Please correct me if i am wrong. Thank you. Bunuel wrote: goodyear2013 wrote: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer? A. 5 B. 7 C. 9 D. 11 E. 13 OE M = 60·61·62·63·64·65·66·67·68·69·70 > 60 = 2·2·3·5 > 62 = 2·31 > 63 = 3·3·7 > 64 = 2·2·17 > 66 = 2·3·11 > 68 = 2·2·17 > 69 = 3·23 > 70 = 2·35 We need to find the power of of 6 in 60*61*62*63*64*65*66*67*68*69*70. 6 = 2*3. Since the power of 2 is higher in 60*61*62*63*64*65*66*67*68*69*70, than the power of 3, then we'll have as many 6's as there are 3's. So, basically we need to find the power of 3 in 60*61*62*63*64*65*66*67*68*69*70: One 3 in 60; Two 3's in 63 (63 = 3^2*7); One 3 in 66; One 3 in 69. Total of five 3's. Therefore the power of 3, as well as power of 6 in 60*61*62*63*64*65*66*67*68*69*70 is 5. Answer: A. Similar questions to practice: whatisthegreatestintegerxforwhich243000000002x140669.htmlwhatisthegreatestintegermforwhichthe163820.htmlifnistheproductofallmultiplesof3between1and101187.htmlifnistheproductofallpositiveintegerslessthan103218.htmlif6yisafactorof102whatisthegreatestpossible129353.htmlfindthenumberoftrailingzerosintheproductof108248.htmlHope it helps.



Math Expert
Joined: 02 Sep 2009
Posts: 47946

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
30 Jun 2014, 02:42
niyantg wrote: Hello Bunuel I tried a different approach. Is it correct ? Since we have to find the number of times 6 occurs between the products of 71 & 59. I Found the number of 3s that come till 70! Minus the number of 3s that come uptill 59!. No of 3s in 70! = 32 & No of 3s till 59! = 27 Subtracting them I got 5, which is the OA. Please correct me if i am wrong. Thank you. Bunuel wrote: goodyear2013 wrote: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer? A. 5 B. 7 C. 9 D. 11 E. 13 OE M = 60·61·62·63·64·65·66·67·68·69·70 > 60 = 2·2·3·5 > 62 = 2·31 > 63 = 3·3·7 > 64 = 2·2·17 > 66 = 2·3·11 > 68 = 2·2·17 > 69 = 3·23 > 70 = 2·35 We need to find the power of of 6 in 60*61*62*63*64*65*66*67*68*69*70. 6 = 2*3. Since the power of 2 is higher in 60*61*62*63*64*65*66*67*68*69*70, than the power of 3, then we'll have as many 6's as there are 3's. So, basically we need to find the power of 3 in 60*61*62*63*64*65*66*67*68*69*70: One 3 in 60; Two 3's in 63 (63 = 3^2*7); One 3 in 66; One 3 in 69. Total of five 3's. Therefore the power of 3, as well as power of 6 in 60*61*62*63*64*65*66*67*68*69*70 is 5. Answer: A. Similar questions to practice: whatisthegreatestintegerxforwhich243000000002x140669.htmlwhatisthegreatestintegermforwhichthe163820.htmlifnistheproductofallmultiplesof3between1and101187.htmlifnistheproductofallpositiveintegerslessthan103218.htmlif6yisafactorof102whatisthegreatestpossible129353.htmlfindthenumberoftrailingzerosintheproductof108248.htmlHope it helps. Yes, that's correct. 60*61*62*63*64*65*66*67*68*69*70 = 70!/59!. The power of 3 in 70!: 70/3 + 70/3^2 + 70/3^3 = 23 + 7 + 2 = 32. The power of 3 in 59!: 59/3 + 59/3^2 + 59/3^3 = 19 + 6 + 2 = 27. The difference = 5.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



VP
Status: It's near  I can see.
Joined: 13 Apr 2013
Posts: 1216
Location: India
Concentration: International Business, Operations
GPA: 3.01
WE: Engineering (Consulting)

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
03 Jul 2015, 10:32
Bunuel wrote: goodyear2013 wrote: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer? A. 5 B. 7 C. 9 D. 11 E. 13 OE M = 60·61·62·63·64·65·66·67·68·69·70 > 60 = 2·2·3·5 > 62 = 2·31 > 63 = 3·3·7 > 64 = 2·2·17 > 66 = 2·3·11 > 68 = 2·2·17 > 69 = 3·23 > 70 = 2·35 We need to find the power of of 6 in 60*61*62*63*64*65*66*67*68*69*70. 6 = 2*3. Since the power of 2 is higher in 60*61*62*63*64*65*66*67*68*69*70, than the power of 3, then we'll have as many 6's as there are 3's. So, basically we need to find the power of 3 in 60*61*62*63*64*65*66*67*68*69*70: One 3 in 60; Two 3's in 63 (63 = 3^2*7); One 3 in 66; One 3 in 69. Total of five 3's. Therefore the power of 3, as well as power of 6 in 60*61*62*63*64*65*66*67*68*69*70 is 5. Answer: A. Similar questions to practice: whatisthegreatestintegerxforwhich243000000002x140669.htmlwhatisthegreatestintegermforwhichthe163820.htmlifnistheproductofallmultiplesof3between1and101187.htmlifnistheproductofallpositiveintegerslessthan103218.htmlif6yisafactorof102whatisthegreatestpossible129353.htmlfindthenumberoftrailingzerosintheproductof108248.htmlHope it helps. I could not understand the highlighted part..why we took only powers of 3 and not of 2. Please help
_________________
"Do not watch clock; Do what it does. KEEP GOING."



Math Expert
Joined: 02 Aug 2009
Posts: 6537

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
03 Jul 2015, 21:47
nailingDcoffin wrote: Bunuel wrote: goodyear2013 wrote: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer? A. 5 B. 7 C. 9 D. 11 E. 13 OE M = 60·61·62·63·64·65·66·67·68·69·70 > 60 = 2·2·3·5 > 62 = 2·31 > 63 = 3·3·7 > 64 = 2·2·17 > 66 = 2·3·11 > 68 = 2·2·17 > 69 = 3·23 > 70 = 2·35 We need to find the power of of 6 in 60*61*62*63*64*65*66*67*68*69*70. 6 = 2*3. Since the power of 2 is higher in 60*61*62*63*64*65*66*67*68*69*70, than the power of 3, then we'll have as many 6's as there are 3's. So, basically we need to find the power of 3 in 60*61*62*63*64*65*66*67*68*69*70: One 3 in 60; Two 3's in 63 (63 = 3^2*7); One 3 in 66; One 3 in 69. Total of five 3's. Therefore the power of 3, as well as power of 6 in 60*61*62*63*64*65*66*67*68*69*70 is 5. Answer: A. Similar questions to practice: whatisthegreatestintegerxforwhich243000000002x140669.htmlwhatisthegreatestintegermforwhichthe163820.htmlifnistheproductofallmultiplesof3between1and101187.htmlifnistheproductofallpositiveintegerslessthan103218.htmlif6yisafactorof102whatisthegreatestpossible129353.htmlfindthenumberoftrailingzerosintheproductof108248.htmlHope it helps. I could not understand the highlighted part..why we took only powers of 3 and not of 2. Please help Hi, we have to take the lower number of power of prime number... if it is a product of consecutive numbers and we are looking for power of 6, which is 2*3.. power of 2 is higher but if you do not have that many number of 3s, how will you make 6 out of it.. so we look at the power of 3 as we know each power of 3 will have a 2 to be multiplied to it to make it 6.. hope it helps
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Intern
Joined: 04 Apr 2015
Posts: 16
Concentration: Human Resources, Healthcare
GMAT Date: 08062015
GPA: 3.83
WE: Editorial and Writing (Journalism and Publishing)

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
04 Jul 2015, 00:49
the product is 60X61X.......70. This can be written as 70!/59! . Now no. of twos are always more than no. of threes, so we need to count no. of threes here. (Similar is the case for even 10s , if we have to find no. of 10s, we must find the no. of 5s present) So, the number of 3s in 70! is 32 and no. of 3s present in 59! is 27. Since we've counted no. of 3s in 59! as well, we must subtract it, hence 3226 = 5



Senior Manager
Status: Countdown Begins...
Joined: 03 Jul 2016
Posts: 310
Location: India
Concentration: Technology, Strategy
GPA: 3.7
WE: Information Technology (Consulting)

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
14 Dec 2016, 11:10
In this we just need to worry about powers of 3.
Following numbers with their power of 3 can be considered: 60  1 63 2 (9*7) 66 1 69  1 Total 5.



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2727

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
29 Jan 2018, 11:06
goodyear2013 wrote: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which \(\frac{M}{6^n}\) is an integer?
A. 5 B. 7 C. 9 D. 11 E. 13 In order for M/6^n to be an integer, we see that M must contain a number of 6’s, all of which will cancel out with the 6^n in the denominator. Since 6 = 2 x 3, we will look for all of the 2and3 pairs contained in the numerator M. We note that there are fewer factors of 3 than factors of 2 contained in M, so we concentrate on finding factors of 3, as follows: 60 has 1 three. 63 has 2 threes. 66 has 1 three. 69 has 1 three. So n is 5. Answer: A
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Manager
Joined: 27 Jul 2017
Posts: 52

Re: If M is the product of all positive integers greater than 59
[#permalink]
Show Tags
03 Apr 2018, 08:18
Bunuel wrote: We need to find the power of 6 in 60*61*62*63*64*65*66*67*68*69*70.
6 = 2*3.
Since the power of 2 is higher in 60*61*62*63*64*65*66*67*68*69*70 than the power of 3, then we'll have as many 6's as there are 3's. So, basically we need to find the power of 3 in 60*61*62*63*64*65*66*67*68*69*70:
One 3 in 60; Two 3's in 63 (63 = 3^2*7); One 3 in 66; One 3 in 69.
Total of five 3's.
Therefore the power of 3, as well as power of 6 in 60*61*62*63*64*65*66*67*68*69*70 is 5.
Thanks Bunuel, though I was able to answer the question correctly, I was solving it by the conventional method and hence took me more time. Your post helped me to learn the best way to solve this type of question.
_________________
Ujjwal Sharing is Gaining!




Re: If M is the product of all positive integers greater than 59 &nbs
[#permalink]
03 Apr 2018, 08:18






