Let n satisfy the condition and n′ be the number obtained by the replacement of digits. Denote by x the number formed by the digits to the left of the replaced digit and denote by y the number formed by the digits to the right of the replaced digit. Let k be the number of digits in y. Then n=x⋅10k+1+4⋅10k+y and n′=x⋅10k+2+22⋅10k+y. As n′ is divisible by n, then so must also be n′−n and 10n−n′, which yields
x⋅10k+1+4⋅10k+y∣9x⋅10k+1+18⋅10k,(2)
x⋅10k+1+4⋅10k+y∣18⋅10k+9y.(3)
We will consider the following cases.
* If x=0, then by (2), t(4⋅10k+y)=18⋅10k for some integer t. As
5(4⋅10k+y)≥5⋅4⋅10k>18⋅10k>3⋅5⋅10k>3(4⋅10k+y),
the only option is t=4. The equation 4(4⋅10k+y)=18⋅10k yields
y=42⋅10k=5⋅10k−1<10k. Altogether n=45⋅10k−1.
* If x=1, then by (2), t(14⋅10k+y)=108⋅10k for some integer t. As
8(14⋅10k+y)≥8⋅14⋅10k>108⋅10k>6⋅15⋅10k>6(14⋅10k+y),
the only option is t=7, but the equation 7(14⋅10k+y)=108⋅10k has no integer solutions, because 7∤108⋅10k.
* If x=2, then by (2), t(24⋅10k+y)=198⋅10k for some integer t. As
9(24⋅10k+y)≥9⋅24⋅10k>198⋅10k>7⋅25⋅10k>7(24⋅10k+y),
the only option is t=8. The equation 8(24⋅10k+y)=198⋅10k yields y=86⋅10k=75⋅10k−2<10k. Altogether n=2475⋅10k−2.
* If x≥3, then by (3) we have x⋅10k+1+4⋅10k+y≤18⋅10k+9y, which yields x⋅10k+1≤14⋅10k+8y. On the other hand x⋅10k+1≥30⋅10k>22⋅10k>14⋅10k+8y. The equations contradict each other, so no such n can exist.