How about 2178?
I started off the problem as follows:
First off the number is in the form of 1000*a + 100*b + 10*c + d. If a is any higher than 2, then when it's multiplied by 4 it won't be 4 digits any more, so a must be either 1 or 2. 10*b + a must be divisible by 4, because the reversed number of 1000*d + 100*c + 10*b + a must be divisible by four, and any number divisible by four has its last two digits divisible by four. That means that a can't be 1 because no multiply of four has a 1 in the ones digit. As a matter of fact, the only multiples of four that are less than 100 with a 2 in the ones place are 12, 32, 52, 72, and 92. 2500 doesn't work out, and when it is multiplied by 4, it becomes 5 digits. Therefore, b must be 1 or 3. The maximum, 2399, when multiplied out is 8796. That means that there will be no carry over from the multiplication of the previous digits, and that since 2*4 = 8, d must be 8. So far, we have a = 2, b = 1 or 3, and d = 8. That leaves 20 different numbers to check. I brute force checked them to find this one.
There are 4 possible answers: 32, 53, 74 and 95.
There are 4 possible answers: 32, 53, 74 and 95.
one
The sum of all palindromic numbers from 1001 to 9999 is 495000.
69
2178 * 4 = 8712
1089! times 9 is 9801. I'm really bored.
2178. 2178 x 4 = 8712
There are 4 possible answers: 32, 53, 74 and 95.
There are 4 possible answers: 32, 53, 74 and 95.
10
next number to this series is 554325. its look like this 15--add 1 to first digit and we get 2 and repeat two times and leave other digit.225 225-- its first digit is 2 and add 1 to it get 3 and write two time and write remaining digit as 3325. 3325-- its first digit is 3 and add 1 to it get 4 and write two time and write remaining digit as 44325. 44325-- its first digit is 4 and add 1 to it get 3 and write two time and write remaining digit as 554325.
9999
4082 Since the tens digit is 2 times the thousand digit, it must be an even digit. So it can be 8, 6, 4, or 2. But, the thousands digit is 4 greater than the hundred digit. So that the hundred digit must be 0, the thousands digit must be 4, the hundreds digit must be 8, and the ones digit must be 2.
The sum of all palindromic numbers from 1001 to 9999 is 495000.
9999 9999
one