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Thursday, April 07, 2005 - 09:36 AM
The numbers 12 and 13 are peculiar...12*12 = 144 but when the digits are reversed the product is also reversed
21*21 = 441
13*13 = 169 but 31*31 = 961
12*13 = 156 but 31*21 = 651
12*14 = 168 but 21*41 = 861
12*22 = 264 but 22*21 = 462
13*22 = 286 but 31*22 = 682
12*23 = 276 but 32*21 = 672
13*23 = 299 but 32*31 = 992
12*24 = 288 but 42*21 = 882
There are more combinations in 112, 113. But I have not tried beyond that. I leave that to the readers to experiment.
Rajan
21*21 = 441
13*13 = 169 but 31*31 = 961
12*13 = 156 but 31*21 = 651
12*14 = 168 but 21*41 = 861
12*22 = 264 but 22*21 = 462
13*22 = 286 but 31*22 = 682
12*23 = 276 but 32*21 = 672
13*23 = 299 but 32*31 = 992
12*24 = 288 but 42*21 = 882
There are more combinations in 112, 113. But I have not tried beyond that. I leave that to the readers to experiment.
Rajan
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Comments
very curious?
here's some more i found!
102*102=10404
201*201=40401
103*103=10609
301*301=90601
122*122=14884
221*221=48841
1022*1022=1044484
2201*2201=4844401
12 x 34 = 408 yet 21 x 43 = 903
13 x 56 = 728 yet 31 x 65 = 2015
and a lot more.....
Is there any limit to using the reversal of digits?
You're right about the numbers 12 and 13. they are very peculiar in an interesting way indeed.
what's more interesting is that different combinations of numbers as beginning with 1 and ending with 2, or 3 yields the same result.
1002*1002 = 1004004
2001*2001 = 4004001
1003*1003 = 1006009
3001*3001 = 9006001
i think the crucial combination is the use of the numbers in the digits between the beginning and ending.
they might work definitely if all in-between digits are zero. other combinations are definitely possible.
(10a+b)(10c+d)=100(ac)+10(ad+bc)+1(bd)
(10b+a)(10d+c)=100(bd)+10(ad+bc)+1(ac)
Do you notice how place values get switched around?
Now all you have to do is hope that the ad+bc part doesn't carry on into the next place value. Which is a pretty safe bet when all of your digits are below 5. However use:
21*89 vs. 12*98
They are no where close together.
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