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12 Jun 2020, 11:44
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B
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The lowest integer that has both positive integers x and y as its factors can be written in the form of \(8^a*9^b*15^c*14^d\). If x can be written as \(2^a*3^{(2b+c)}*5^c*7^{(d-1)}\), where a, b, c and d are positive integers, what are the possible number of values that y can take ?
A. \((2b+c+1)d\)
B. \((2b+c+1)(c+1)\)
C. \((a+1)(b+1)(c+1)(d+1)\)
D. \((a+ 1)(2b+c+1)(c+1)d\)
E. \((3a+ d + 1)(2b+c+1)(c+1)(d +1)\)
Are You Up For the Challenge: 700 Level Questions
A. \((2b+c+1)d\)
B. \((2b+c+1)(c+1)\)
C. \((a+1)(b+1)(c+1)(d+1)\)
D. \((a+ 1)(2b+c+1)(c+1)d\)
E. \((3a+ d + 1)(2b+c+1)(c+1)(d +1)\)
Are You Up For the Challenge: 700 Level Questions
12 Jun 2020, 16:14
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IMO:- B
LCM (x, y) = 8^a∗9^b∗15^c∗14^d
or LCM (x,y)= 2^(3a+d) * 3^ (2b+c) * 5^c * 7^d----------------------(1)
x= 2^a * 3^ (2b+c) * 5^c * 7 ^(d-1)-------------(2)
Any LCM will take the highest powers of the prime factors present in x and y.
IMP Point-- Firstly we need to compare power of prime factors with the following criteria:-
>if the LCM contains the power of a prime factor greater which is present in x, then that power of that prime factor must be present in y.
>While if the LCM contains the power of a prime factor equal to that in x, then y can take any power of that prime factor not greater than the power of the prime factor present in the LCM.
Using equation (1) & (2)
Comparing LCM with x we can say
For Y, 2 will only be in form 2^ 3a+d ( as Power is higher in LCM in comparison to x)---------------(3)
Similarly, For Y, 7 will only be in form 7^d ( as Power is higher in LCM in comparison to x)-------------(4)
While the power of 3 is equal in LCM and x, hence For Y 3 can take values as 3^0, 3^1, 3^2......3^2b+c
or Total values Y can take for the power of 3 will be (2b+c+1)----as zero is included-----------------------(5)
Similarly the power of 5 is equal in LCM and x, hence For Y 5 can take values as 5^0, 5^1, 5^2......5^c
or Total values Y can take for the power of 3 will be (c+1)----as zero is included-----------------------(6)
Combining all the above 4 ways
Total Possible way = 1* 1* (2b+c+1) * (c+1) (Only 1 way for 2 & 7)
Hope solution is clear
LCM (x, y) = 8^a∗9^b∗15^c∗14^d
or LCM (x,y)= 2^(3a+d) * 3^ (2b+c) * 5^c * 7^d----------------------(1)
x= 2^a * 3^ (2b+c) * 5^c * 7 ^(d-1)-------------(2)
Any LCM will take the highest powers of the prime factors present in x and y.
IMP Point-- Firstly we need to compare power of prime factors with the following criteria:-
>if the LCM contains the power of a prime factor greater which is present in x, then that power of that prime factor must be present in y.
>While if the LCM contains the power of a prime factor equal to that in x, then y can take any power of that prime factor not greater than the power of the prime factor present in the LCM.
Using equation (1) & (2)
Comparing LCM with x we can say
For Y, 2 will only be in form 2^ 3a+d ( as Power is higher in LCM in comparison to x)---------------(3)
Similarly, For Y, 7 will only be in form 7^d ( as Power is higher in LCM in comparison to x)-------------(4)
While the power of 3 is equal in LCM and x, hence For Y 3 can take values as 3^0, 3^1, 3^2......3^2b+c
or Total values Y can take for the power of 3 will be (2b+c+1)----as zero is included-----------------------(5)
Similarly the power of 5 is equal in LCM and x, hence For Y 5 can take values as 5^0, 5^1, 5^2......5^c
or Total values Y can take for the power of 3 will be (c+1)----as zero is included-----------------------(6)
Combining all the above 4 ways
Total Possible way = 1* 1* (2b+c+1) * (c+1) (Only 1 way for 2 & 7)
Hope solution is clear
General Discussion
13 Jun 2020, 05:55
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Bunuel
Rewriting the LCM as a proper prime factorization, we find
\(\\
\text{LCM(}x, y\text{)} = (8^a)(9^b)(15^c)(14^d) = (2^3)^a (3^2)^b (3 \times 5)^c (2 \times 7)^d = (2^{3a}) (3^{2b}) (3^c)(5^c)(2^d)(7^d) = (2^{3a + d} )(3^{2b + c})(5^c)(7^d)\\
\)
If this is the LCM of x and y, then y must be a divisor of this number, so the only prime factors of y can be 2, 3, 5 and 7.
We know
\(\\
x = (2^a)(3^{2b + c})(5^c)(7^{d-1})\\
\)
When we find the LCM of x and y, for each prime base, we'll choose the larger exponent. So in the prime factorization of y, we must have precisely \(2^{3a + d}\) and \(7^d\), because those exponents appear in the LCM but smaller exponents appear in the prime factorization of x. The exponents on 3 and 5, however, are identical in the prime factorization of x and in the LCM, so x is giving us all the 3's and 5's that we need. So in the prime factorization of y, as long as the exponents on the "3" and on the "5" are less than or equal to 2b + c and c, respectively, we'll get the correct exponents on "3" and on "5" when we find the LCM. So in the prime factorization of y, we have 2b + c + 1 options for the exponent on "3" (we add one because the exponent can be zero), and similarly c + 1 options for the exponent on the "5". Multiplying our choices, there are (2b + c + 1)(c + 1) possibilities in total, and B is the answer.
