Eind the smallest number which leaves remainders
8 and 12 when divided by 28 and 32
respectively
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Answered by
2
Answer:
Given that,
The smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively.
28-8=20 and 32-12=20 are divisible by the required numbers.
Therefore the required number will be 20 less than the L.C.M of 28 and 32.
Now,
PRIME FACTORISATION OF 28
= 2 × 2 × 7
PRIME FACTORISATION OF 32
= 2 × 2 × 2 × 2 × 2
L.C.M of 28 and 32
= 2×2×2×2×2×7
= 224
THEREFORE THE REQUIRED SMALLEST NUMBER IS 224-20 = 204.
HOPE IT HELPED..
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