find the smallest number that divided by 45 55 and 120 leaving remainder 5
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Firstly subtract 5 from the numbers
that is 45-5=40
55-5=50
120-5=115
Now, by using euclid's division algorithm
115=50×2+15
50=15×3+5
15=5×3+0
So HCF (115,50) =5
Now,
40=5×8+0
So, HCF (115,50,40)=5
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that is 45-5=40
55-5=50
120-5=115
Now, by using euclid's division algorithm
115=50×2+15
50=15×3+5
15=5×3+0
So HCF (115,50) =5
Now,
40=5×8+0
So, HCF (115,50,40)=5
I HOPE IT WILL HELP YOU
PLEASE MARK AS BRAINLIEST ANSWER
shaksham2:
you are brilliant
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