Monday Madness

How's about a little simple math? What is the smallest number divisible by 13 which, divided by any of the numbers from 2 to 12 inclusive, leaves a remainder of 1?

Too easy.

Answer is 83,161

This is a problem in modular algebra
a mod b is defined as a integer divisable by b with a remainder of a

let N be the number of interest
N=0 mod 13
N=1 mod 2
N=1 mod 3
…….
N=1 mod 12

If a number is mod 2, it is also mod 4
if a number is mod 3 it is also mod 6
if a number is mod 4 it is also mod 8
if a number is mod 5 it is also mod 10
if a number is mod 6 it is also mod 12
then our problem reduces to

N=0 mod 13
N=1 mod 7 , 1 mod 8, 1 mod 9, 1 mod 10, 1 mod 11, 1 mod 12
but if a number is mod 8 and mod 12 then it is also mod 24

N=0 mod 13
N=1 mod 7, 1 mod 9, 1 mod 10, 1 mod 11, 1 mod 24
if a number is mod 10 and mod 24 then it is also mod 120
but a number mod 7 and 9 ithen it is also mod 63

N=0 mod 13
N= 1 mod 63 , 1 mod 11 , 1 mod 120
if a number is mod 63, 11 and 120 then it is also mod [63x11x120] or mod 83,160

N=0 mod 13
1 mod 83,160

N= 1+83,160
N=1+2[83160]
etc
Is 83161= 0 mod 13 YES
is 183 = 1 mod J 2 83,161 answer

http://www.freemathhelp.com/forum/threads/…emainder-1-when

Well done Lee!

83,161 is the correct answer.

Modular algebra is just a "different" way of looking at least common multiplier (which is how I solved it). [(LCM)n+1]/13 = whole number. In this case, LCM = 27,720. n=3.

Anyhow… here's your treat, oatmeal raisin cookies:

[external image: Posted Image]

Are they low GI (I'm a diabetic)? Never mind, my wife will devour them