Back with another question around a 4 digit number (ABCD) - again the digits can repeat and not necessarily distinct
1. AB + CD results in a number which is part of an unique number series
2. This number satisfies the condition X ^ Y where X and Y are positive integers
3. This number is a product of two numbers C and D and in turn if we add C + D it again results in a number which is part of a special series
4. C and D can also be expressed as X1 ^ Y1 and X2 ^ Y2
4. DCBA (reverse of this number) + a perfect square results in another well known number!
Find ABCD
1. AB + CD results in a number which is part of an unique number series
2. This number satisfies the condition X ^ Y where X and Y are positive integers
3. This number is a product of two numbers C and D and in turn if we add C + D it again results in a number which is part of a special series
4. C and D can also be expressed as X1 ^ Y1 and X2 ^ Y2
4. DCBA (reverse of this number) + a perfect square results in another well known number!
Find ABCD
Answer to the above puzzle...
ReplyDeleteABCD is 9261 and here's how
1. 92 + 61 = 153 which is the first Armstrong Number
2. 9261 = 21 ^ 3
3. 9261 = 343 x 27. 343 + 27 = 370 which is again an Armstrong Number
4. 343 = 7 ^ 3 and 27 = 3 ^ 3
5. 1629 is reverse of 9261 and if you add 100 it results in 1729 which is Ramanujam Number