
Topic review (newest first)
 ganesh
 20060322 17:35:46
I think the proof that it is not possible for single digit numbers can be extrapolated for multiple digit numbers also. I will post my reply after examining all the possibilities.
 sunfish
 20060322 17:27:48
Thanks.
But what I really have problem figuring out is ...what if A or B is >9? ie. A and B can be numbers with more than 1 digit....AxB=AB possible?
 ganesh
 20060322 12:10:49
I have tried and now I have got a proof that there cannot be a number of any number of digits ABCD such that ABCD=AxBxCxD. I shall first prove for three digit numbers.
If AxBxC=ABC, then, either all the three are equal, or one or two of them are not equal.
First, lets consider the case of either one or two of A,B,C are not equal. That is all the three are not equal.
The smallest cannot be zero, since then the product would be zero.
The smallest cannot be one, since then, even if the other two numbers take the maximum possible value, the product is only 81, not a three digit number.
The smallest number cannot be 2, since, even if the other two numbers take the maximum value, that is 9 and 9, the product is only 162, and a number greater than 200 is not obtained.
The smallest number cannot be 3 since, even if the other two numbers take the maximum value, that is 9 and 9, the prouct is less than 300.    Similarly, the smallest number cannot be 8, since even if the other two numbers take the maximum value, that is 9 and 9, the product is less than 800.
Now, the only option is the three numbers are equal. When they are, that is, 9, 9, and 9, the product is only 729, less than 900.
Thus, it is seen that the smallest number cannot be 0,1,2,3,4,5,6,7,8, or 9. Therefore, there can be no such three digit number.
Extrapolating to any digit of n numbers, there can be no number ABCD such that AxBxCxD=ABCD.
 ganesh
 20060321 19:15:19
sunfish, if AxBxC=ABC, then, 100A+10B+C=ABC. 100A+10B=ABCC=C(AB1) Therefore, C=(100A+10B)/(AB1). We would have to try out 9x9=81 combinations for A=1,2,3,4,5,6,7,8, or 9 and B=1,2,3,4,5,6,7,8, or 9. None of the numbers, A, B or C can be zero, as the produce would then be zero. Let us explore other possibilities to know whether such numbers exist or not. Just give me some time.
 sunfish
 20060321 18:56:19
Thanks for the reply:) yup I tried that.
What if AB, A and B are numbers with more than 1 digit? Any possibility?
for example X=123, Y=12, XY=12312
 ganesh
 20060321 18:09:53
sunfish, Your question is, (if I get you right!), is there a two digit number AB such that A x B = AB, correct? That is, AB should be equal to 10A+B. AB=10A+B ABB=10A B(A1)=10A B = 10A/(A1). Value of A Value of B 1 Infinity 2 20 3 15 4 40/3 5 25/2 6 12 7 35/3 8 80/7 9 45/4 0 0
It can be seen that for single digit numbers of A, we don't get single digit numbers of B with the exception of 0,0. Hence, there can be no such number
 sunfish
 20060321 17:36:41
Hi, need help with this...Is it possible to have a number AB whereby A multiply B = AB? how do I work this out?
Note : AB is a number where by the front portion is made up of A and the rear is made up of B...if you get what I mean...
for eg. if X=2 and Y=3, then the number XY is 23
Thanks
