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Determine 3-digit numbers abc which divided by 7,8 and 9 give the remainder 1,4 and 7.

 Nov 1, 2019
 #1
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Divisibility rules of 8 and 9 are easiest, so let's find that first.

 

Modulo may be a bit hard for now, so I'll show you a more rudimentary way.

 

The list of numbers that when divided by 9 give a remainder of 7 are 7, 16, 25, 34, 43, 52

Same for 8 that give 4: 4, 12, 20, 28, 36, 44, 52


We see that 52 meets both of these requirements! This will happen every cycle of 72, because the LCM of 8 and 9 is 72.

(in other words, 52, 124, 196, 268, 340, ...)

Now, the three digit numbers are 124, 196, 268, etc.

 

We still haven't used the divisible by 7 with remainder of 1 rule!

Those numbers are 1, 8, 15, 22, etc.

If we test every single number that we had before (124, 196, 268, 340, 412, 484, 556, 628, 772, 854, 916, 988), we can find that

484 can be divided by 7 to give a remainder of 1. So can 988.

 

Therefore, there are two numbers - 484 and 988.

There is a faster way to do this - with modulo, but try looking at that for yourself! Search up "modulo" online and start looking at the math sites.

 

You are very welcome!

:P

 Nov 1, 2019
 #2
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+1

n mod 7 = 1, n mod 8 = 4, n mod 9 = 7, solve for n

 

Using "Chinese Remainder Theorem + Modular Multiplicative Inverse", we get:

 

n = 504m + 484, where m = 0, 1, 2, 3.......etc. 

 Nov 1, 2019

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