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If the sum of the digits in the odd positions starting from the ones digit less the sum of the digits in the even positions (starting with the 10s digit) is divisible by 11, then so is the original number.

for 8003:

odd position digits: 3 + 0 = 3

even position digits: 0 + 8 = 8

3 - 8 = -5 which is not divisible by 11, so 8003 is NOT divisible by 11.

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βˆ™ 7y ago
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βˆ™ 7y ago

8003 ÷ 11 = 727 remainder 6

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Q: IS 8003 divisible by 11
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