Question
Find the largest number which $615$ and $963$ leaving remainder $6$ in each case.
✓
Answer
We have to find the largest number which divides $(615 - 6)$ and $(963 - 6)$ exactly.
Therefore, the required number $= HCF$ of $609$ and $957$ Resolving $609$ and $957$ into prime factors,
we have: $609 = 3 \times 7 \times 29 957 = 3 \times 11 \times 29$
Therefore, $HCF$ of $609$ and $957$ $= 29 \times 3 = 87$
Hence, the required largest number is $87.$
Therefore, the required number $= HCF$ of $609$ and $957$ Resolving $609$ and $957$ into prime factors,
we have: $609 = 3 \times 7 \times 29 957 = 3 \times 11 \times 29$
Therefore, $HCF$ of $609$ and $957$ $= 29 \times 3 = 87$
Hence, the required largest number is $87.$
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