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Let the four-digit number be xyzw. This number when multiplied by 9 should give a four-digit number which is wzyx. Now the maximum possible four-digit number is 9999. Therefore, xyzw has to be less or equal to 1111 because if, for instance, the number is 1112, then upon multiplication by 9, we get 10008 which is a five-digit number.
The smallest four-digit number possible is 1000.
Therefore, the number has to be between 1000 and 1111. Any number between 1000 and 1111 will have 1 as the first digit. So, one thing is certain: the first digit is 1; which implies x = 1. Thus, the number is of the form: 1yzw.
When 1yzw is multiplied by 9, the result should be wzy1. This is possible only if w = 9; 9 x 9 = 81 which gives 1 in the unit's place and 8 is carried over. Thus, w = 9, and the number is of the form 1yz9.
Now, the task is to find 'y' and 'z'. Now, let's consider the step where 'y' is multiplied by 9. In this step, 9 x y should give a single-digit number because if we get a double-digit number, it will mean a carry over. If there is a carry over at this stage, then the final result will be a five-digit number and not a four-digit one. Getting messy isn't it? Let me illustrate with an example: The number is of the form 1yz9; let z = 0 (to keep things simple) and let y = 2. Then the number we have is: 1209. Let's multiply by 9; when we come to the step where we multiply 2 by 9 we get 18 which requires 1 to be carried over; we, then multiply 1 by 9 and add 1 (the carry over). This gives us 10 and the resulting number is a five-digit number (1209 x 9 = 10881) and not a four-digit one.
It is only when 'y' is either 0 or 1 that there will be no carry over. Thus, y = 0 or y = 1. So, the number is of the form 10z9 or 11z9.
Well, let's consider the number to be 11z9 and not 10z9. In this case, when 'z' is equal or greater than 1, we get a five-digit number on multiplying by 9. For instance, the lowest possible number in this case is 1119 and the product of 1119 and 9 is 10071 - a five-digit number. So, 'z' has to be 0 and the number we finally arrive at is 1109. But the product of 1109 and 9 is 9981 which is not the reverse of 1109. Thus, 11z9 is not the correct choice. The correct choice has to be 10z9.
Phew! I have made things messier and messier. But, now, I can see some light. Since the number is 10z9, the product of 10z9 and 9 has to be 9z01. So, now, let's multiply 10z9 by 9. In the first step we have 9 x 9 = 81; that leaves me 1 in the ones place and 8 is carried over. In the second step we multiply 'z' by 9 and add 8. This gives 9z + 8. This should give 0 in the tens place of the result (9z01). This is possible only when z = 8; in this case we have 9z + 8 = (9x8) + 8 = 80. Thus, we have 0 in the tens place and 8 is carried over. The number we finally arrive at is, therefore, 1089. The product of 1089 and 9 is 9801, which is indeed the reverse of 1089.
I have solved this in the most messy manner. I am sure there must be a simpler way of solving it. Please let me know of a simpler solution.
Some useful links for
- Union Public Service Commission - www.upsc.gov.in
- IIT-Kharagpur - www.iitkgp.ac.in
- Indian Statistical Institute - www.isical.ac.in
- Indian Institute of Technology Madras - www.iitm.ac.in
- Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
- Indian Institute of Mass Commission - www.iimc.nic.in
- IIT Bombay - www.iitb.ac.in
- Indian School of Mines, Dhanbad - www.ismdhanbad.ac.in
- Birla Institute of Technology, Ranchi - www.bitmesra.ac.in
- Central Institute of Fisheries Nautical and Engineering Training - www.cifnet.nic.in
- Indian Institute of Information Technology, Allahabad (Deemed University) - www.iiita.ac.in
- Central Marine Fisheries Research Institute, Kochi - www.cmfri.com
- Tata Institute of Social Sciences, Mumbai - www.tiss.edu