Number Theory

AdkAdi

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Joined
Jun 14, 2021
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49
I have figured out that a and d have 6 as a common factor but I am not able to figure out what to do nextIMG_20210917_133506_184.jpg
 

iamsteve03

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Oct 19, 2021
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First things first, this is a MUST BE kind of a question on the concept of GCD(HCF); therefore, there is bound to be more than one case. As such, instead of trying to write down cases, a better approach would be to use concepts of GCD to solve the question.

What makes this question doubly challenging is the fact that we do not have values of any of the numbers and hence, we will have to rely on the procedure of finding out the HCF using prime factorization method, to bail us out.

Let’s have a look at what data we have on hand. We know that,

GCD(a,b) = 24 = 2323 * 3. Therefore, we can conclude that both a and b should have 2323 and 3 in them. In other words, they should be a multiple of 24.

GCD (b,c) = 36 = 2222 * 3232. So, b and c should be multiples of 36. Additionally, b should be a multiple of 72, since it has to be a multiple of 24 also.

GCD (c,d) = 54 = 2∗332∗33. So, c and d should be multiples of 54. Additionally, c should be a multiple of 108 since it has to be a multiple of 54 also.

It’s mentioned that the GCD(d,a) has to be in between 70 and 100. Also, from the above discussion, we know that the GCD(d,a) will be 2121 * 3131. This is because, we always take the lowest powers of the prime factors while calculating the GCD.

But 2121 * 3131 is 6, which is not between 70 and 100. In order for this to be in the specified range, we need to look for a number, which when multiplied with 2121 * 3131 will give us a number in this range. Clearly, this has to be 13, since 13*6 = 78. You can also see that, if we multiply any of the other numbers by 6, they will either be less than 70 or more than 100.

This means that, a MUST be divisible by 13 in order that the gcd(d,a) is in between 70 and 100. So, the correct answer option is D.

As an alternative, you can take suitable values for a, b, c and d and work out the problem in similar manner.

Hope this helps!
 

AdkAdi

New member
Joined
Jun 14, 2021
Messages
49
First things first, this is a MUST BE kind of a question on the concept of GCD(HCF); therefore, there is bound to be more than one case. As such, instead of trying to write down cases, a better approach would be to use concepts of GCD to solve the question.

What makes this question doubly challenging is the fact that we do not have values of any of the numbers and hence, we will have to rely on the procedure of finding out the HCF using prime factorization method, to bail us out.

Let’s have a look at what data we have on hand. We know that,

GCD(a,b) = 24 = 2323 * 3. Therefore, we can conclude that both a and b should have 2323 and 3 in them. In other words, they should be a multiple of 24.

GCD (b,c) = 36 = 2222 * 3232. So, b and c should be multiples of 36. Additionally, b should be a multiple of 72, since it has to be a multiple of 24 also.

GCD (c,d) = 54 = 2∗332∗33. So, c and d should be multiples of 54. Additionally, c should be a multiple of 108 since it has to be a multiple of 54 also.

It’s mentioned that the GCD(d,a) has to be in between 70 and 100. Also, from the above discussion, we know that the GCD(d,a) will be 2121 * 3131. This is because, we always take the lowest powers of the prime factors while calculating the GCD.

But 2121 * 3131 is 6, which is not between 70 and 100. In order for this to be in the specified range, we need to look for a number, which when multiplied with 2121 * 3131 will give us a number in this range. Clearly, this has to be 13, since 13*6 = 78. You can also see that, if we multiply any of the other numbers by 6, they will either be less than 70 or more than 100.

This means that, a MUST be divisible by 13 in order that the gcd(d,a) is in between 70 and 100. So, the correct answer option is D.

As an alternative, you can take suitable values for a, b, c and d and work out the problem in similar manner.

Hope this helps!
Thank you very much
 
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