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If M is the product of all positive integers greater than 59 [#permalink]
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29 Jun 2014, 02:44
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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
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Re: If M is the product of all positive integers greater than 59 [#permalink]
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Re: If M is the product of all positive integers greater than 59 [#permalink]
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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.



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Re: If M is the product of all positive integers greater than 59 [#permalink]
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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.
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Re: If M is the product of all positive integers greater than 59 [#permalink]
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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
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Re: If M is the product of all positive integers greater than 59 [#permalink]
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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
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Re: If M is the product of all positive integers greater than 59 [#permalink]
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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



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Re: If M is the product of all positive integers greater than 59 [#permalink]
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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.
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Re: If M is the product of all positive integers greater than 59 [#permalink]
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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
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Re: If M is the product of all positive integers greater than 59 [#permalink]
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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.
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