Another one

How many positive four-digit integers n have the property that the three-digit number obtained by removing the leftmost digit is one-ninth of n?

n=1000a+100b+10c+d                    where a,b,c and d are all single digit numbers.

n-1000a = n/9

9n-9000a=n

8n=9000a

8000a+800b+80c+8d=9000a

800b+80c+8d=1000a

100b+10c+d=125a

If a=1       

100b+10c+d=125       so the number is   1125

if a=2

100b+10c+d=250       so the number is    2250

if a=3

100b+10c+d=375       so the number is    3375

if a=4

100b+10c+d=500       so the number is    4500

if a=5

100b+10c+d=625       so the number is    5625

if a=6

100b+10c+d=750       so the number is    6750

if a=7

100b+10c+d=875       so the number is    7875

if a=8

100b+10c+d=1000       so the number is    9000    That is no good

if a=9

100b+10c+d=1125       so the number is  5 digits which is too big.

So there are exactly 7 four digit numbers that meet this description.

Melody: I found only 7 such numbers:

n=1107; m=1;if((n - 1000*m) / n ==1/9, goto3, goto4);printn;n=n+9; if(n<10000, goto2, discard=0;

n = 1125, 2250, 3375, 4500, 5625, 6750, 7875

Yes sorry.

It has just been pointed out to me that I mad a careless error that trivialized the answer. (My wrong answer I mean.)

It should not have been 8, it should have been 8d in the working.

I'll fix that and then I will think about it