Amazing number

Since the given condition must apply when $n = 10$ , the digit sum of the original two-digit number must be $10$ , (since if the two-digit number is $ab$ , then $10 * ab = ab0$ ).

This narrows the options to $19, 28, 37, 46, 55, 64, 73, 82$ and $91$ . Now multiply each of these numbers by $n = 6$ and keep those for which the digit sum of the product is equal to $6$ . This leaves us with $19, 37$ and $55$ as our options. Now do the same with $n = 7$ , We are left with just $19$ . We can then quickly check that the given condition is satisfied for $6 \le n \le 15$ , confirming that $\boxed{19}$ is indeed the desired two-digit number.

int n,k,p,h,j,z,x,sum,q;
for(n=10;n<=99;n++) {
for(k=6;k<=15;k++){ p=n*k; h=p%10; p/=10; j=p%10; p/=10; z=p%10; p/=10; x=p%10; sum=x+z+j+h; if(sum==k) { System.out.println(""+n+" "+k); }

well, just keep in mind that when (ab).(xy), the sum total of the product will be (a+b).(x+y). so, we have to find the two digit no. which when divided by n gives a product whose sum of the digits is n.

Therefore, (a*b).(n) should give n as the sum of the digits of the product obtained. so "(a+b)" must be equal to 10 as (1+0)=1 and when 1 multiplied by n gives n as the sum total.

Thus the required numbers can be 19, 28, 37, 46, 55, 64, 73, 82, 91. but in all cases except 19, the sum total of the digits is done more than once. for ex: (28).(15) [taking n=15] =420 =6 which is equal to 15 as 1+5 = 4+2 =6 [taking into account the sum of the digits]

but in case of 19. (19).(15) =285 =(2+8+5) =15

hence, 19 is the correct answer....