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Beast 3C p.61 #121


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Haha! K.I.S.S. I'm completely shaking my head & laughing. :). He was attempting to use an 'alternate method' discussed by the book. 54/53=R1, 55/53=R2 SO (54x55)/53 must be the same as (1x2)/53.


Instead of looking closely & suggesting the more traditional method I just became locked onto that. Methinks I'll go make coffee now. :leaving:

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I don't have the book but I suspect they want you to solve this by using the property that

the remainder of the division of a product (in this case 54x55) with a number (53) is the

product of the remainders of the division of the 2 factors of the product (54 and 55) with that

number (53).



r 54/53 = 1,

r 55/53 = 2


r (54x55)/53 = 1x2 = 2

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Here's the shortcut way BA is approaching it:




Since we are only looking for remainders, and not the actual product, they separate it out to find the remainders of 54/53 and 55/53


54/53 has a remainder of 1

55/53 has a remainder of 2


Since we were originally multiplying 54x55, we have to multiply their remainders 2x1=2

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Ugh. I finally get it. DH came home & explained it in, like, 1second. It's simpler than I was making it. 2 isn't divisible by 53. That's where I got stuck. "well,then,dear,it's ZERO. Remainder 2.". All while giving me his most southern bless your heart expression. I don't know why I couldn't produce zero. Thanks for y'all's help.

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