Find LCM of 18 and 24 .
LCM is appearently 240 .
Divisors :
40_{divisor}={1,2,4,5,8,10,20,40} ;
120_{divisor}={1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120} ;
140_{divisor}={1,2,4,5,7,10,14,20,28,35,70,140}
Multiples :
40_{multiple}={40,80,120,160,200,240,280,320,360,400,440,480,520,560,600,640,680,720,760,800,840,...} ;
120_{multiple}={120,240,360,480,600,720,840,960,...} ;
140_{multiple}={140,280,420,560,700,840,980,...} .
Green rows are for GCD .
GCD of the numbers given = ?
Have a look at ry1.
A=2x3x ... x 19 and B=1x2x3x ... x14x15. This means A = 16x17x18x19 x B , i.e. a multiple of B.
16x17=16x(16+1)=256+16=272 ... and the question : 272x342 = ?
A=2x3x ... x53x54 and B=1x2x3x ... x54x55
There are multiples of 5 but only a lone 5 in each set since the sets are of prime factors .
The product of 3 numbers is not the product of their LCM and GCD .
The product of 2 numbers is the product of their LCM and GCD.
It should be a prime factor, isn't it ?
LCM(50,78) = ?
The sum GCD(150,160) + LCM(150,160) = ?
The first couple of elements of the ordered primes set: 2,3,5,7,11,13,17,19,23,29,31,37,...
LCM(4,9,13) = ?
LCM(79,83) = ?
LCM(12,16) = ?
mod-26 practically means divide by 26 and get the residue. Have please a look at ry1.
Treat letters as if a "prime factor" since their values are not known .
The first is 3xB times the second, i.e. a multiple .
The denominator has GCD = 4 and LCM = 2640 .
Have please a look at ry1 .
An LCM question. 5 hours are enough till their destination .
LCM(10, 12 , 14) = ? .
The numbers are almost huge but the method stays the same .
This is the feedback!