Question 407951
The solution of x+5=2(MOD7) and x is an integer between 1 and 12,then how many values does x have
<pre><font face = "batangche" color = "indigo" size = 4><b>
x + 5 = 2(mod 7)

Replace 2(mod 7) by 7y + 2

x + 5 = 7y + 2

x = 7y - 3

Since 

{{{1 <= x <= 12}}}

Replace x by 7y - 3

{{{1 <= 7y - 3 <= 12}}}

Add 3 to all 3 sides

{{{4 <= 7y <= 15}}}

Divide all three sides by 7

{{{4/7<=y<=15/7}}}

{{{4/7<=y<=2&1/7}}}

Since y is an integer,

{{{1<=y<=2}}}

So either y = 1 or y = 2

Two solutions. 

When y = 1, x = 7y - 3 = 7(1) - 3 = 4
When y = 2, x = 7y - 3 = 7(2) - 3 = 14 - 3 = 11

Edwin</pre>