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# Modular Help!

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Is 21^100-12^100 a multiple of 11? Thanks!

Feb 24, 2019

#1
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We have two ways to present this solution, one by logic.

Solution 1: Take the bases for the start. Note that 21 is one less than a multiple of 11, namely 22, so we have 21 is congruent to -1(mod 11), and 12 is congruent to 1(mod 11). Thus, we can rewrite it as $$(-1)^{100}-1^{100}$$ , and that is congruent to   $$1-1$$ or 0(mod 11), so yes it is a multiple of 11.

Solution 2: Since 21 is congruent to 10(mod 11) and 12 is congruent to 1(mod 11), the value is congruent to $$10^{100}-1^{100}=10^{100}-1$$ mod 11.

This is where you have to use some logic. $$10^{100}$$ contains 101 digits (100 zeroes and one 1). By subtracting one, one hundred 9's would be remaining, and by the divisibility rule of 11, we have to add the alternating digits and subtract them, which would then yield 0(mod 11). So, it is divisible.

There you go.

Feb 24, 2019
#2
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$$21^{100}-12^{100} = (3\cdot 7)^{100} - (3\cdot 4)^{100} = \\ 3^{100}(7^{100}-4^{100})\\ \text{if you factor the second term all the way down you will have a factor of }\\ (7+4)=11\\ \text{and thus this number is a multiple of }11$$

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Feb 24, 2019