there is two digit number the ten's place is exceeds the unit place by 6,and when 8 is added to the number, the number reversed itself,
what is the number??
2006-06-19 00:56:26 · 4 answers · asked by manjul c 1 in Science & Mathematics ➔ Mathematics
The first number has to be (x+6) and the second number is (x). There are four possible combinations: 6,0 (60); 7,1 (71); 8,2 (82); 9,3 (93). 60 plus 8 (68) does not equal 06, or 6. 71 plus 8 does not equal 17. 82 plus 8 does not equal 28. 93 plus 8 does not equal 39.
Unfortunately, this problem is impossible.
Sorry I couldn't be of more help.
2006-06-19 16:38:44 · answer #1 · answered by Remember 9-11 2 · 0⤊ 0⤋
60 + 8 = 68 (ie reversed 86, not 60)
71 + 8 = 79 (ie reversed 97, not 17)
82 + 8 = 90 (ie reversed 9, not 28)
93 + 8 = 101 (ie reversed 101, not 39)
not possible, sorry!
2006-06-19 00:59:59 · answer #2 · answered by aus_sie_girl 2 · 1⤊ 0⤋
This is impossible:
Let the two digit number be ab=10a+b
If ab+8=10a+b+8=ba=10b+a, then 9b=9a+8. This is impossible because 9 divides 9b, and 9 divides 9a, but 9 does not divide 8.
It can't be done.
It doesn't even matter that aâ¥b+6
2006-06-19 01:02:29 · answer #3 · answered by Eulercrosser 4 · 1⤊ 0⤋
Using my amazing brain capacity I declare this question impossible!
2006-06-19 03:42:32 · answer #4 · answered by quickster94 3 · 1⤊ 1⤋