find the largest no that divides 231 and 156 leaving a remainder 6 in each case
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We have to find the greatest number which divides 6 1 5 and 9 6 3 leaving a remainder of 6 in every case.
So let's subtract 6 from 6 1 5 and 9 6 3
6 1 5-6=6 0 9
9 6 3-6=9 5 7
Now let's find HCF
Prime factorization of 6 0 9=2 9×3×3
Prime factorization of 9 5 7=2 9×3×1 1
Let us now find the common factors in both the cases
2 9 and 3
x=2 9×3=8 7
Step-by-step explanation:
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