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The greatest common divisor of two integers is x + 5 and their least common multiple is x(x + 5) , where x is a positive integer. If one of the integers is 60, what is the smallest possible value of the other one?

 Feb 14, 2022
 #1
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a=1; b=1;d=GCD(a, 60) ;e=LCM(a, 60); if(d==b+5 and e==b*(b +5), goto5, goto6);printd,e, a, b; a++;if(a<500, goto2, 0);a=1;b++;if(b<500, goto2, discard=0; 

 

GCD of [a,  60] ==x + 5

LCM of [a, 60]==x * [x + 5]

 

a ==100

x == 15

 

GCD[100, 60] ==20 ==(x + 5)

LCM[100, 60]==300==x * (x + 5)

 Feb 14, 2022
 #2
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I’d set x+5=y, so x=y−5 and so we know that y should be a divisor of 60 greater than 5. Hence we can only have:

 

y∈{6,10,12,15,20,30,60}

 

On the other hand, we need y(y−5) to be a multiple of 60. We can make a table:

 

y         y−5     y(y−5)

6          1            6

10        5           50

12        7          84

15       10        150

20      15        300

30      25       750

60      55       3,300

 

and we see that only two cases are possible, namely y=20 or y=60.

Now that we know that, if N is the other number, we need:

60N=y^2(y−5)

For y=20 we get N=100

For y=60 we get N=3300

 

Since we need the smallest positive integers, we therfore have:

x ==15, and the other number=100

GCD(60, 100)= 20

LCM(60, 100)==300

 Feb 14, 2022
edited by Guest  Feb 14, 2022

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