Question 175629This question is from textbook Algebra 1
: If the digits of a two-digit positive interger are reversed, the result is 6 less than twice the original numder. Find all such intergers for which this is true.
How do you figure this out?
This question is from textbook Algebra 1
Found 4 solutions by ankor@dixie-net.com, Mathtut, josmiceli, solver91311:
Answer by ankor@dixie-net.com(22740)   (Show Source):
Answer by Mathtut(3670)  (Show Source):
You can put this solution on YOUR website! lets call our digits a and b written as ab
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now remember ab can also be written as 10a+b
and when it is reverse it is ba which can be written as 10b+a
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so with that info lets write the equation: they only gave us enough info for one equation so we are going to have to do some surmising on this one:
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10b+a=2(10a+b)-6
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10b+a=20a+2b-6
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8b-19a=-6
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the only possibilities for this scenario is digits 0-9 for both a and b. the question is can we find an integer value for both a and b that satisfies this equation.
the value of 19a always has to be more that the value of 8b in order to get -6
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so for a=1 b can only be 1 or 2 (8-19),(16-19)...neither of which works
a=2 so 19a=38.. so we need 8b=32...8(4)=32 so we have a pair a=2 b=4.
a=3 so 19a=57...so we need 8b=51....no integer value will work
a=4 so 19a=76...so we need 8b=70....no integer value will work
a=5 so 19a=95...so we need 8b=89---> we have exhaused our possibilities here. as you can see in order for 8b=89 we would have to multiply by two digit number and as the value of 19a gets higher the same will hold true
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so we have found the only integers for which this is true and
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they are
Answer by josmiceli(19441)  (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website! Let the tens digit be represented by  and the ones digit be represented by  . That means you can represent any two digit positive number by  if you restrict and to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
If our original number is then the number with reversed digits is  . Then the conditions of the problem give us:
Simplify and solve for
Now remember that both and are restricted to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. That means that we have to find which of the given set of numbers, when substituted for , yield also an element of the given set. We know that since the denominator is 19 and there are only ten consecutive integer possibilities, we have at most 1 solution.
We can also see that  must be an integer multiple of 19.
The first five integer multiples of 19 are 19, 38, 57, 76, and 95. So:
 →  → not an integer.
 →  →  IS an integer. This is our solution, but let's continue just to make sure.
 →  → not an integer.
 →  → not an integer.
 →  → not an integer AND  so we can stop looking.
Now we know the ones digit is 4, so using  we know the tens digit is 2.
The only positive two-digit integer that satisfies the given conditions is 24.
Check the answer:
2 times 24 is 48 which is 6 more than 42 which is 24 with the digits reversed. Checks.
By the way, the word is integer, not interger.
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