Question
Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
✓
Answer
We need to find the largest number which divides 615 and 963 leaving remainder 6 in each case.
The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.
Therefore,
The required number = H.C.F. of 609 and 957.
By applying Euclid’s division lemma
957 = 609 × 1 + 348
609 = 348 × 1 + 261
348 = 216 × 1 + 87
261 = 87 × 3 + 0.
Therefore, H.C.F. = 87.
Hence, the required number is 87.
The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.
Therefore,
The required number = H.C.F. of 609 and 957.
By applying Euclid’s division lemma
957 = 609 × 1 + 348
609 = 348 × 1 + 261
348 = 216 × 1 + 87
261 = 87 × 3 + 0.
Therefore, H.C.F. = 87.
Hence, the required number is 87.
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