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JoeM12
Hi, why also not 130?
realloggervick
The highest common factor and least common multiple of two positive integers are 13 and 520 respectively. If one number lies between 90 and 150, what is the value of that number?

(A) 91
(B) 104
(C) 117
(D) 130
(E) 143
Since LCM of two numbers is \(2^3*5*13\), if you take one number as \(13*5*2\), the other number will need to have \(2^3\) => and since we are given that HCF is 13, if the other number has more 2's too, the HCF changes to 26. Therefore, we can't take 130.
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where did you get " LCM of two numbers is \(2^3*5*13\)"? I'm missing something
Krunaal
JoeM12
Hi, why also not 130?
realloggervick
The highest common factor and least common multiple of two positive integers are 13 and 520 respectively. If one number lies between 90 and 150, what is the value of that number?

(A) 91
(B) 104
(C) 117
(D) 130
(E) 143
Since LCM of two numbers is \(2^3*5*13\), if you take one number as \(13*5*2\), the other number will need to have \(2^3\) => and since we are given that HCF is 13, if the other number has more 2's too, the HCF changes to 26. Therefore, we can't take 130.
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mastergmatster
where did you get " LCM of two numbers is \(2^3*5*13\)"? I'm missing something
Krunaal
JoeM12
Hi, why also not 130?
Since LCM of two numbers is \(2^3*5*13\), if you take one number as \(13*5*2\), the other number will need to have \(2^3\) => and since we are given that HCF is 13, if the other number has more 2's too, the HCF changes to 26. Therefore, we can't take 130.

”The highest common factor and least common multiple of two positive integers are 13 and 520 respectively.”
\(520 = 2^3*5*13\)
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realloggervick
The highest common factor and least common multiple of two positive integers are 13 and 520 respectively. If one number lies between 90 and 150, what is the value of that number?

(A) 91
(B) 104
(C) 117
(D) 130
(E) 143
HCF X LCM = Product of two numbers
13 X 520 = 13a X 13b
6760 = 169ab
ab = 40

(1,40) OR (5,8)

(1,40) -> The numbers are 13 and 520
(5,8) -> The numbers are 65 and 104

104 lies between 90 and 150

Therefore B

Please note that -
For (4,10) numbers -> (52, 130)
and (2,20) numbers -> (26, 260)
The LCM of each of these pairs is not 520, it is 260
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Hi,
IN the below, the product 40 can also be achieved by (4X10) which would be equivalent to 130x52. How do we rule this out?

PSKhore
realloggervick
The highest common factor and least common multiple of two positive integers are 13 and 520 respectively. If one number lies between 90 and 150, what is the value of that number?

(A) 91
(B) 104
(C) 117
(D) 130
(E) 143
HCF X LCM = Product of two numbers
13 X 520 = 13a X 13b
6760 = 169ab
ab = 40

(1,40) OR (5,8)

(1,40) -> The numbers are 13 and 520
(5,8) -> The numbers are 65 and 104

104 lies between 90 and 150

Therefore B
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Sarthak_shukla
Hi,
IN the below, the product 40 can also be achieved by (4X10) which would be equivalent to 130x52. How do we rule this out?

PSKhore
realloggervick
The highest common factor and least common multiple of two positive integers are 13 and 520 respectively. If one number lies between 90 and 150, what is the value of that number?

(A) 91
(B) 104
(C) 117
(D) 130
(E) 143
HCF X LCM = Product of two numbers
13 X 520 = 13a X 13b
6760 = 169ab
ab = 40

(1,40) OR (5,8)

(1,40) -> The numbers are 13 and 520
(5,8) -> The numbers are 65 and 104

104 lies between 90 and 150

Therefore B
Hi,

For (4,10) numbers -> (52, 130)
and (2,20) numbers -> (26, 260)
The LCM of each of these pairs is not 520, it is 260
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Is it mandatory in the steps above to group similar numbers together .
For example ( 2^3*5) = 40 so we took one number as 13*2^8 , 13*5 = ( 104 , 65 ) ?
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