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MCII-B Number Theory 4.26
How would I solve this problem? I looked at the corresponding example problem, Example 4.6, and I'm still sort of lost...what should I do?
Remember that the divisibility rule for 11 can basically extend to a "remainder" rule: Any number (mod 11) is equivalent to it's alternating sum of digits (right to left). So, for example, $52 \equiv 2-5 = -3 \pmod{11}$ and $522\equiv 2-2+5 = 5 \pmod{11}$. This should help.
Wait, for 5222, would the remainder be (2+2)-(5+2), or (5+2)-(2+2)? I added the 1st and 3rd digits together, and also added the 2nd and 4th digits...
For (mod 11), I would recommend getting use to alternating + and - from right to left. We're really using the rule proved in 4.25b (which helps understand why it's right to left).
I'd like to let you figure out the problem (and it's a bad example for explaining the rule because of the repeated digits), so let's look at a few other examples:
- $1234 \equiv +4-3+2-1 = 2 \pmod{11}$
- $9753 \equiv +3-5+7-9 = -4 \equiv 7 \pmod{11}$
- $97531 \equiv +1-3+5-7+9 = 5 \pmod{11}$
Hope this helps.
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