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How many numbers that are not divisible by 6 divide evenly [#permalink]
 28 Jan 2012, 01:09
Question Stats:
27% (02:56) correct
72% (01:30) wrong based on 11 sessions
How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72 Any idea how to solve this please?
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Last edited by Bunuel on 28 Jan 2012, 02:09, edited 1 time in total.
Added the OA
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Re: Numbers divisible by 6 [#permalink]
28 Jan 2012, 02:08
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enigma123 wrote: How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
Any idea how to solve this please? Finding the Number of Factors of an IntegerFirst make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers. The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors. For more on number properties check: math-number-theory-88376.htmlBACK TO THE ORIGINAL QUESTION: How many numbers that are not divisible by 6 divide evenly into 264,600?(A) 9 (B) 36 (C) 51 (D) 63 (E) 72 264,600=2^3*3^3*5^2*7^2, thus it has total of (3+1)(3+1)(2+1)(2+1)=144 differernt positive factors, including 1 and the number itself. # of factors that ARE divisible by 6 will be 3*3*(2+1)(2+1)=81: we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting. So, # of factors that ARE NOT divisible by 6 is 144-81=63. Answer: D. Another solution here: new-set-of-good-ps-85440.html#p642384Hope now it's clear.
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Re: How many numbers that are not divisible by 6 divide evenly [#permalink]
28 Jan 2012, 15:59
Thanks Bunuel - but in the solution that's there in the link which is presented by Atish, I am struggling to understand why he did Now if we add the two numbers above we end up double counting the factors of 5^2*7^2 = (2+1)*(2+1) = 9Can you please explain?
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Re: How many numbers that are not divisible by 6 divide evenly [#permalink]
28 Jan 2012, 16:22
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enigma123 wrote: Thanks Bunuel - but in the solution that's there in the link which is presented by Atish, I am struggling to understand why he did
Now if we add the two numbers above we end up double counting the factors of 5^2*7^2 = (2+1)*(2+1) = 9
Can you please explain? This part is from another approach (direct counting), which is in my solution there too. Maybe it will answer your question: How many numbers that are not divisible by 6 divide evenly into 264,600? 264,600=2^3*3^3*5^2*7^2We should find the factor which contain no 2 and 3 together, so not to be divisible by 6. Clearly, the factors which contain only 2, 5, 7 and 3, 5, 7 won't be divisible by 6. So how many such factors are there? 2^3*5^2*7^2 --> (3+1)*(2+1)*(2+1)=36; 3^3*5^2*7^2 --> (3+1)*(2+1)*(2+1)=36; 36+36=72. Here comes the part you have a problem with. This number (72) contains duplicates, (some factors which are not divisible by 6 are counted twice): both 36'es count the factors which have ONLY 5's and/or 7's. (5*7=35, 5*7^2=245, 5^2*7=175, 5*7^0=5, 5^0*7=7....), so basically factors of 5^2*7^2 are counted twice. How, many such factors does 5^2*7^2 have? (2+1)*(2+1)=9. So we should subtract this 9 duplicated factors from 72 --> 72-9=63. Hope it's clear.
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Re: How many numbers that are not divisible by 6 divide evenly [#permalink]
28 Jan 2012, 16:29
Perfect and many thanks.
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Re: Numbers divisible by 6 [#permalink]
15 Sep 2012, 14:36
Bunuel wrote: enigma123 wrote: How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
Any idea how to solve this please? Finding the Number of Factors of an IntegerFirst make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers. The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors. For more on number properties check: math-number-theory-88376.htmlBACK TO THE ORIGINAL QUESTION: How many numbers that are not divisible by 6 divide evenly into 264,600?(A) 9 (B) 36 (C) 51 (D) 63 (E) 72 264,600=2^3*3^3*5^2*7^2, thus it has total of (3+1)(3+1)(2+1)(2+1)=144 differernt positive factors, including 1 and the number itself. # of factors that ARE divisible by 6 will be 3*3*(2+1)(2+1)=81: we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting. So, # of factors that ARE NOT divisible by 6 is 144-81=63. Answer: D. Another solution here: new-set-of-good-ps-85440.html#p642384Hope now it's clear. Hey Bunuel, So if I was interested in knowing the number of factors 264,600=2^3*3^3*5^2*7^2 that are divisible by 35=5*7 Answer: (3+1)(3+1)(2)(2)=64 Your saying I shouldn't add 1 to powers the primes 5 and 7? How about how many factors of 264,600=2^3*3^3*5^2*7^2 are divisible by 3675=3*(5^2)*(7^2)? Would it be (3+1)(3)(2)(2)=48?
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Re: Numbers divisible by 6 [#permalink]
15 Sep 2012, 16:23
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alphabeta1234 wrote: Bunuel wrote: enigma123 wrote: How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
Any idea how to solve this please? Finding the Number of Factors of an IntegerFirst make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers. The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors. For more on number properties check: math-number-theory-88376.htmlBACK TO THE ORIGINAL QUESTION: How many numbers that are not divisible by 6 divide evenly into 264,600?(A) 9 (B) 36 (C) 51 (D) 63 (E) 72 264,600=2^3*3^3*5^2*7^2, thus it has total of (3+1)(3+1)(2+1)(2+1)=144 differernt positive factors, including 1 and the number itself. # of factors that ARE divisible by 6 will be 3*3*(2+1)(2+1)=81: we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting. So, # of factors that ARE NOT divisible by 6 is 144-81=63. Answer: D. Another solution here: new-set-of-good-ps-85440.html#p642384Hope now it's clear. Hey Bunuel, So if I was interested in knowing the number of factors 264,600=2^3*3^3*5^2*7^2 that are divisible by 35=5*7 Answer: (3+1)(3+1)(2)(2)=64 Your saying I shouldn't add 1 to powers the primes 5 and 7? How about how many factors of 264,600=2^3*3^3*5^2*7^2 are divisible by 3675=3*(5^2)*(7^2)? Would it be (3+1)(3)(2)(2)=48?NO. It should be (3+1)(3)(1)(1)=12. 2 can be at any power between 0 and 3; 3 can be at any power between 1 and 3 - we need at least a factor of 3; 5 and 7 are already at power 2 in 3675, so just one choice for each. You have to ensure that you have each prime factor of 3675 at least at the power it shows in the factorization of 3675, but not greater than the power of that factor in the decomposition of 264,600.
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Re: Numbers divisible by 6 [#permalink]
15 Sep 2012, 17:19
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NO.
It should be (3+1)(3)(1)(1)=12.
2 can be at any power between 0 and 3;
3 can be at any power between 1 and 3 - we need at least a factor of 3;
5 and 7 are already at power 2 in 3675, so just one choice for each.
You have to ensure that you have each prime factor of 3675 at least at the power it shows in the factorization of 3675, but not greater than the power of that factor in the decomposition of 264,600.[/quote]
EvaJager,
Correct me if I am wrong but is this the method your using: How many factors of 264,600=2^3*3^3*5^2*7^2 are divisible by 3675=3*(5^2)*(7^2)?
You are essentially: 264,600/3675=(2^3*3^3*5^2*7^2)/(3*5^2*7^2)=2^3*3^2*5^0*7^0. Now we find the number of factors of 2^3*3^2*5^0*7^0 which is (3+1)(2+1)(0+1)(0+1). Do I have your method down correctly? Is this what your doing in your head??
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Re: Numbers divisible by 6 [#permalink]
16 Sep 2012, 00:24
alphabeta1234 wrote: NO.
It should be (3+1)(3)(1)(1)=12.
2 can be at any power between 0 and 3;
3 can be at any power between 1 and 3 - we need at least a factor of 3;
5 and 7 are already at power 2 in 3675, so just one choice for each.
You have to ensure that you have each prime factor of 3675 at least at the power it shows in the factorization of 3675, but not greater than the power of that factor in the decomposition of 264,600. EvaJager, Correct me if I am wrong but is this the method your using: How many factors of 264,600=2^3*3^3*5^2*7^2 are divisible by 3675=3*(5^2)*(7^2)? You are essentially: 264,600/3675=(2^3*3^3*5^2*7^2)/(3*5^2*7^2)=2^3*3^2*5^0*7^0. Now we find the number of factors of 2^3*3^2*5^0*7^0 which is (3+1)(2+1)(0+1)(0+1). Do I have your method down correctly? Is this what your doing in your head??Yes, exactly!
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How many numbers that are not divisible by 6 divide evenly i [#permalink]
22 Mar 2013, 13:07
How many numbers that are not divisible by 6 divide evenly into 264,600?
9
72
51
63
36
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Re: How many numbers that are not divisible by 6 divide evenly i [#permalink]
22 Mar 2013, 13:11
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Re: How many numbers that are not divisible by 6 divide evenly [#permalink]
09 May 2013, 10:31
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How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
264600 = (2^3) * (3^3) * (5^2) * (7^2)
Numbers not divisible by 6 ----->(factors that are not multiples of 3) + (factors that are not multiples of 2) - (factors that are multiples of neither 2 nor 3)
( number of factors of 2^3 * 5^2 *7^2 ) + ( number of factors of 3^3 * 5^2 *7^2 ) - ( number of factors of 5^2 *7^2 )
= [(3+1)(2+1)(2+1)] + [(3+1)(2+1)(2+1)] - (2+1)(2+1)]
=36 + 36 -9 = 63
Note : (Subtract number of factors of 5^2 *7^2 , because you have counted them twice .)
HTH
-Jyothi
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Re: How many numbers that are not divisible by 6 divide evenly
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09 May 2013, 10:31
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