SOLUTION: How many two-digit integers are increased by 11 when the order of the digits is reversed?

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Question 677032: How many two-digit integers are increased by 11 when the order of the digits is reversed?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If the units digit is +a+ and
the tens digit is +b+, the the
value of the number is +a+%2B+10b+
given:
+a+%2B+10b+=+b+%2B+10a+%2B+11+
+%28+10b+-+b+%29+-+%28+10a+-+a+%29+=+11+
+%28+10+-+1+%29%2Ab+-+%28+10+-+1+%29%2Aa+=+11+
+9b+-+9a+=+11+
+9%2A%28+b+-+a+%29+=+11+
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I don't think it's true for any 2-digit numbers
+b+-+a+ must be a whole number and not
a fraction, so
+b+-+a+=+%28+9k+%29+%2F+9+ where +k+ can
be 0 to 9 . Note that if +k+=+10+, then
+%28+9%2A10+%29+%2F+9+=+10+, but the difference of
2 digits can't be +10+
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and also, +9k+ can never be +11+