Question

Find the number in which remainder
dovided by 7, 11, 13. is, 1, 2, 3​

Answer 1

Step-by-step explanation:

this is yr answer

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Answer 2

Step-by-step explanation:

Let no. be n

n=7a+1

n=9b+2

n=11c+3

Equations are solvable since only integral solutions are allowed.

11c+3=9b+2

11c+1=9b

b=(11c+1)÷9

put c=1,2...10.at c=4, b=5.

c=13, b=16..... so on

Now look at

11c+3=7a+1

a=(11c+2)÷7

At c=3,a=5

c=10,a=16.... so on.

from first equation, c=9k+4

From second set c=7m +3.

or our first equation

9k+4=7m+3

m=(9k+1)÷7

k=3,m=4....so on

for the lowest possible no.

k=3,m=4

c=31

n=11×31 +3=344.