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01 Dec 2021, 08:15
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
45% (02:58) correct
55%
(02:26)
wrong
based on 69
sessions
History
Date
Time
Result
Not Attempted Yet
For some positive integer n, the number \(110n^3\) has 110 positive integer divisors, including 1 and the number \(110n^3\). How many positive integer divisors does the number \(81n^4\) have?
(A) 110
(B) 191
(C) 261
(D) 325
(E) 425
(A) 110
(B) 191
(C) 261
(D) 325
(E) 425
01 Dec 2021, 09:19
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In this question we know that 110 * \(n^3\) has 110 factors.
110 can be prime factorized as 5 * 2 * 11. So the powers of these numbers will play a role in deciding the total number of factors.
We can first start by breaking down the factors into possible combinations -
110 can be formed either by 11 * 10 or 11 * 2 * 5 or 55 * 2.
Now we know that the total number of positive factors for any number in the prime factorized form is given by
\(x^p\) * \(y^q\) ... = (p+1) * (q+1) ...
Lets back track the given number of factors the above information -
Few points to consider -
1) As \(n^3\) is present, each unit of prime factorized n will now have the power multiplied by 3. So the each power of the prime number that constitutes n will now be thrice.
For example -
If N = \(x^p\) * \(y^q\)
\(N^3\) = \(x^{3*p}\) * \(y^{3*q}\)
Each of the power will be a multiple of 3.
2) If we see the constituents of 110, we will realize that the possible total number of factors of \(n^3\) can be 54 ( 55 - 1)
Thus the number 110 * \(n^3\) is possibly formed by (54 + 1) * (1 + 1)
Also, if this combination were true, n definitely consists of two of the three prime factors among 11, 2 and 5 (it doesn't matter what are they) as long as can find the possible power.
Lets say, n has a combination of 2 and 5
N = \(2 ^ x\) * \(3 ^ y\) * \(p ^ z\)
\(N ^ 3\) = \(2 ^ 3^x\) * \(5 ^ 3^y\) * \(p ^ 3^z\)
110 * \(N ^ 3\) = 11 * 5 * 2 * \(2 ^ 3^x\) * \(5^ 3^y\) * \(p ^ 3^z\)
Total number of factors of 110 * \(N ^ 3\) = (1 + 1) * ( 3x + 1 + 1) * (3 y + 1 + 1) * ( 3z + 1) = 2 * 5 * 11
Simplifying further -
(2) * ( 3x + 2) * (3 y + 2) * ( 3z + 1) = 2 * ( 3 * 1 + 2 ) * ( 3 * 3 + 2)
Now we can see that for the equation to satisfy, z = 0 and x = 1 & y = 3
Thus, we have found out what N can possibly be N = \(2^1\) * \(5^3\)
Note: This is just one possible value of N.
Lets verify if the number of factor match -
Number of factors of =
= 110 * (\(2^1\) * \(5^3\)) * (\(2^1\) * \(5^3\)) * (\(2^1\) * \(5^3\))
= 2 * 5 * 11 * (\(2^1\) * \(5^3\)) * (\(2^1\) * \(5^3\)) * (\(2^1\) * \(5^3\))
= 11 * \(2 ^ 4\) * \(5 ^ {10}\)
Total factors = 2 * 5 * 11 = 100
If you are with me till here, the question asks us to find the number of factors of 81 * \(n^4\)
81 * \(n^4\) = \(3 ^ 4\) * \(2 ^ {1*4} \) * \(2 ^ {3*4}\)
81 * \(n^4\) = \(3 ^ 4\) * \(2 ^ 4 \) * \(2 ^ {12} \)
Total factors = (4+1) * (4+1) * (12+1) = 5 * 5 * 13
= 325
IMO - D
110 can be prime factorized as 5 * 2 * 11. So the powers of these numbers will play a role in deciding the total number of factors.
We can first start by breaking down the factors into possible combinations -
110 can be formed either by 11 * 10 or 11 * 2 * 5 or 55 * 2.
Now we know that the total number of positive factors for any number in the prime factorized form is given by
\(x^p\) * \(y^q\) ... = (p+1) * (q+1) ...
Lets back track the given number of factors the above information -
Few points to consider -
1) As \(n^3\) is present, each unit of prime factorized n will now have the power multiplied by 3. So the each power of the prime number that constitutes n will now be thrice.
For example -
If N = \(x^p\) * \(y^q\)
\(N^3\) = \(x^{3*p}\) * \(y^{3*q}\)
Each of the power will be a multiple of 3.
2) If we see the constituents of 110, we will realize that the possible total number of factors of \(n^3\) can be 54 ( 55 - 1)
Thus the number 110 * \(n^3\) is possibly formed by (54 + 1) * (1 + 1)
Also, if this combination were true, n definitely consists of two of the three prime factors among 11, 2 and 5 (it doesn't matter what are they) as long as can find the possible power.
Lets say, n has a combination of 2 and 5
N = \(2 ^ x\) * \(3 ^ y\) * \(p ^ z\)
\(N ^ 3\) = \(2 ^ 3^x\) * \(5 ^ 3^y\) * \(p ^ 3^z\)
110 * \(N ^ 3\) = 11 * 5 * 2 * \(2 ^ 3^x\) * \(5^ 3^y\) * \(p ^ 3^z\)
Total number of factors of 110 * \(N ^ 3\) = (1 + 1) * ( 3x + 1 + 1) * (3 y + 1 + 1) * ( 3z + 1) = 2 * 5 * 11
Simplifying further -
(2) * ( 3x + 2) * (3 y + 2) * ( 3z + 1) = 2 * ( 3 * 1 + 2 ) * ( 3 * 3 + 2)
Now we can see that for the equation to satisfy, z = 0 and x = 1 & y = 3
Thus, we have found out what N can possibly be N = \(2^1\) * \(5^3\)
Note: This is just one possible value of N.
Lets verify if the number of factor match -
Number of factors of =
= 110 * (\(2^1\) * \(5^3\)) * (\(2^1\) * \(5^3\)) * (\(2^1\) * \(5^3\))
= 2 * 5 * 11 * (\(2^1\) * \(5^3\)) * (\(2^1\) * \(5^3\)) * (\(2^1\) * \(5^3\))
= 11 * \(2 ^ 4\) * \(5 ^ {10}\)
Total factors = 2 * 5 * 11 = 100
If you are with me till here, the question asks us to find the number of factors of 81 * \(n^4\)
81 * \(n^4\) = \(3 ^ 4\) * \(2 ^ {1*4} \) * \(2 ^ {3*4}\)
81 * \(n^4\) = \(3 ^ 4\) * \(2 ^ 4 \) * \(2 ^ {12} \)
Total factors = (4+1) * (4+1) * (12+1) = 5 * 5 * 13
= 325
IMO - D
