Find the least number which is divisible by 1 2 6 , 1 6 8 and 1 4 7 perfectly (without leaving any remainder)
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Nice coloring of numbers, haha XD
Nice solution...
How did you color the text of your solution?
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Use L A T E X codes as follows....
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Thanks mate!
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@Anik Mandal – So now, reply to my comment in the new thing you learnt, reply me in a C O L O R F U L way :D
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@Aditya Raut – A n i k M a n d a l
A d i t y a R a u t
Learnt to write colourfully! How's that? :)
2 * 7square is 98 !!
The answer is the LCM of 126,168 and 147,which is 3 5 2 8
find the LCM of 126, 168, and 147. That's all
a=126;b=168;c=147; r = (a b)/gcd(a,b); r = r c/gcd(r,c); and r is the answer
........126,..168,..147
7.....|..18,.....24,....21
7.....|..18,.....24,....3
3.....|...18,....24,....1
3.....|.....6,......8
2.....|....2,.......8,
1.....|....1,........4,
7 * 7 * 3 * 3 * 2 * 4 = 3528
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The answer is just the LCM of 1 2 6 , 1 6 8 and 1 4 7 , and
1 2 6 = 2 × 7 × 3 2
1 6 8 = 2 3 × 3 × 7
1 4 7 = 2 × 7 2
Hence the LCM is 2 3 × 3 2 × 7 2 = 8 × 9 × 4 9 = 3 5 2 8