LCM and HCF Reunion

Number Theory Level 4

How many unordered pairs of integers $(h_1,h_2)$ exist such that their lowest common multiple (LCM) is 1729 and their highest common factor (HCF) is 91?

4 18 0 9 2 None of these choices 1

2 solutions

Sandeep Bhardwaj
Jul 9, 2015

$\begin{cases} 91=13 \times 7 \\ 1729=13 \times 7 \times 19 \end{cases}$ All the un-ordered pairs $(h_1,h_2)$ are $(91,1729),(-91,1729),(91,-1729)$ and $(-91,-1729)$ .

Hence total number of un-ordered pairs is $4$ .

Common mistake:

Missed to include the negative numbers.

enjoy!

Moderator note:

This problem deals with the simple case where $LCM = HCF \times Prime$ .

How can you generalize this to the product of more primes?

How can you determine the lcm of (-91), 1729????

Ankit Kumar Jain - 5 years, 10 months ago

Hariharan Gandhi
Jul 10, 2015

=>HCF * LCM=a * b

=>Let a=h1

=>Let b=h2

=>LCM=1729=(13)x (133)

        =(13)x(7)x(19)

=>HCF=91=(13)x(7)

=>HCF*LCM=(13^2)x(7^2)x(19)=(h1)x(h2)

=>The powers to which 13,7 and 19 are raised are 2,2 and 1 respectively.

=>Hence,no.of possible values for (h1,h2) is (2)x(2)=4 [(h1,h2)is an unordered pair]

Moderator note:

No. Just because $h_1 \times h_2 = 13^2 \times 7^2 \times 19$ does not imply that $\gcd (h_1 , h_ ) = 13 \times 7$ .

You found a necessary, but not sufficient, condition.

LCM and HCF Reunion

2 solutions

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