Math, asked by aneeshappu93, 7 months ago

find the least number which when divided by5,6,8,9 and 12 leaves a reamainder1 in each case but when divided by 13 leaves no remainder

Answers

Answered by vnk0148

1

Answer:

3601

Step-by-step explanation:

LCM of 5, 6, 8, 9 and 12=2×2×2×3×3×5=360

∴Requirednumber=(360k+1)

(360k+1) is divisible by 13 then k=10

Required number =360×10+1=3601

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