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Question 149192: A two-digit number is three less than seven times the sum of its digits. If the digits are reversed, the new number is 18 less than the original number. What is the original number?
Answer by 24HoursTutor.com(40)   (Show Source):
You can put this solution on YOUR website! Let the two digits be x and y. Now assume that the digit x is on the tens place value and y on the units place value. Our number would therefore be :
10x+y
Now, adding 3 to this number gives us seven times the sum of the digits(x+y). Hence we get the equation :
10x+y+3=7(x+y)
Now, lets try and get the value of x from this equation in terms of y :
10x+y+3=7x+7y
10x-7x=7y-y-3
3x=6y-3 {Now divide each number by 3}
x=2y-1
Now, the question also stated that we need to reverse the digits, which means that now y will be on the tenths place and x on the units place. Our newnumber would therefore be : 10y+x
Now, according to the given information this number is 18 less than the previous number. Which means if we add 18 to this new number we will get the old number. Our equation therefore looks like :
10y+x+18=10x+y {substituting the value of x derived earlier}
10y+(2y-1)+18=10(2y-1)+y
10y+2y-1+18=20y-10+y
12y-y-20y=-10-17
-9y=-27
9y=27
y=27/9 = 3
Now substituting it to get the value of x=2y-2=2(3)-1=6-1=5
Now we have x=5
y=3
so our number is 10x+y=10(5)+3=50+3=53
You can check and see that all the condition are held to be true!!!
The above question is solved by one of the experts from 24HoursTutor.com
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