|
Question 843199: Find the largest two digit integer that is increased by 75% when it's digits are reversed.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let your digits be a and b.
your first number is ab.
your second number is ba.
the value of your first number is 10a + b
the value of your second number is 10b + a.
assume that your second number is 75% bigger than your first number.
this means that your second number is 1.75 times as big as your first number.
since the value of your first number is 10a + b and the value of your second number is 10b + a, this means that:
10b + a is equal to 1.75 * (10a + b)
simplify this equation to get:
10b + a = 17.5 * a + 1.75 * b
subtract 1.75 * b from both sides of this equation and subtract a from both sides of this equation to get:
10b - 1.75b = 17.5a - a
simplify this to get:
8.25b = 16.5a
divide both sides of this equation by 8.5 to get:
b = 16.5a/8.25
this simplifies to b = 2a
since b has to be an integer, and b has to be 2a, then the biggest that b can be is 8.
if it were 9, than a would not be an integer.
it can't be 10 since a single digit can only go up to 9.
so the biggest that b can be is 8 and this means that the biggest that a can be is 4 which is 1/2 * b.
since your first number is ab, then that number is 48.
since your second number is ba, then that number is 84.
84 is 1.75 * 48 so the solution is good.
|
|
|
| |
