13 mod 2436
Step 1: Usual Euclidean algorithm
2436 = 187*13+5 ---- 1
13=2*5+3 ---- 2
5=1*3+2 ---- 3
3=1*2+1 ---- 4
Step2: Using method of back substitution
From eq 4;
1= 3-1.2
Subs eq 3
1= 3-1.(5-1.3) = 2.3-1.5
Subs eq 2
1=2.(13-2.5)-1.5
1= 2.13-4.5-1.5
1=2.13-5.5
Sub eq 1
1=2.13-5.(2436-187.13)
1=2.13-5.2436+935.13
1=937.13-5.2436
13(937)-2436(5) = 1
13 mod 2346 is 937
