2026/076) Find multiple of 29 having last 2 digits 47

Because last 2 digits are 47 so the number is of the form 100n+47

Because it is multiple of 29 so it is of the form 29m

So 29m = 100n + 47

Or $29m - 100n = 47\cdots(1)$

Because GCD(29,100) =1 this has got  solution

Let us find 1 in form of 100a + 29b

We have $100 = 29*3 + 13\cdots(2)$

$29 = 2 * 13 + 3\cdots(3)$

$13 =  3 *4+1 \cdots(4)$

From (4)

$1 = 13 - 3 * 4$

$= 13 - 4(29 - 2 *13) = 9 * 13 - 4 * 29$  (from (3))

$= 9*(100-3*29) - 4* 29$ (from (2))

$= 9 *100 - 31 *29\cdots(5)$

 

From (1) and (5) we get

 

$47 = 29 + 18$

 

Or $47 = 29 +  18(9 * 100 - 31 *29)$

 

Or $47 = 29 + 100*(162)  - 29 *558$

 

$= 100 *162 - 29 * 557$

 

$-29 * 557 = 47 - 100 *162$

 

Here we need to make the multiple of 100 positive this shall make multiple of 29 positive

 

We need to add 29 * 100 * 6 (600 is least multiple of 100 above 557 ) to LHS and 29* 6 * 100 on RHS both 17400 to get  

 

$29 * (600 - 557) = 47 + 100 *( 29 * 6 - 162$

 

Or $29 * 43 = 100 * 12 + 47= 1247$

 

This is the smallest multiple of 29 having last 2 digits 47 and we have

 

$29 * (43 +100n) = (12 + 29n) *100 + 47 = 1247 + 2900n$ are all positive numbers when $n >=0$