LU,
4.
Find the least number which when divided by 3, 4, 5, 6, 10 and 12 leaves a remainder 2 in each case.
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Question:
Find the least number which when divided by 3, 4, 5, 6, 10 and 12 leaves a remainder 2 in each case.
Answer:
LCM of 3, 5, 6, 8, 10 and 12=2×2×2×3×5=120
∴Requirednumber=(120k+2)
(120k+2) is divisible by 13 then k=8
Required number =120×8+2=962
#Hopeithelps
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