Question

Find the least number which when
successively divided by 2,3 and 7 leavesremainders 1, 2 and 3 respectively​

Answer 1

Answer:

Let us solve it by applying Modular Airthmatic.

Let the Number be X.

X=1 (mod 2) ———— (i)

X=2 (mod 3)————— ( ii)

X=3 (mod 7) ————— (iii)

(i) can be rewritten as ,

X= 1 + 2 t where t is any integer. ————(*)

Putting this X value against (ii),we get

1+2 t =2 (mod 3)

=> 2 t=1 (mod 3) ————— (iv)

In order to eliminate 2 from 2t, we have to divide 2 t by 2. But in Modular Airthmatic, except some special cases, direct division as is usual in classical Airthmatic ,is not allowed and in stead it should be multiplied by MODULAR INVERSE.

[ 3=2 (1)+1 : 1=3 (1) + 2 (-1) : -1 (mod 3)=2

So 2 is the Modular Inverse 2 modulo 3]

Multiplying (iv) both the sides by 2 ,we get

t = 2 (mod 3)

=> t=2 +3 u where u is any integer.

Putting the value t in (*) , we get

X=1 + 2 (2+3u)= 5 +6 u —————(v)

Putting the value of X against (iii) above ,we get

5+6u=3 (mod 7) ———— (vi)

6u = -2 (mod 7)=5 (mod 7) —— (vii)

6 is the Modular Inverse 6 Modulo 7

Multiplying (vii ) by 6 both sides, we get

u =30 (mod 7) = 2 (mod 7)

=> u = 2 +7 v where v is any integer.

Putting this u value in (v) above, we get

X=5+6 (2+7v) =5+12+42 v

=> X=17 +42 v

For Smallest value of X,

Take v=0

X=17 : The Smallest Positive value. □ANSWER.

[ Take v= -1 , X= -25 : The Largest Negative value. There are Infinity negative values & Positive values ]