A two digit numbers is three times the sum of the digits. It is 45 less than the number formed when the digits are interchanged. Find the two digits
Please anyone to help me with solution
let AB be the number. Then
10A+B=3(A+B) or 7A-2B=0
10B+A-45=10A+B or 9A-9B=-45
7A-2B=0
9A-9B=-45 Now, solving that (multipy second by 7 and first by 9)
63A-18B=0
63A-63B=-315
subtract second eq from first.
-18B+63B=315
45B=315
B=7
you solve for A
Let the number be written as ab
Then its value is 10a+b
So, now we are told that
10a+b = 3(a+b)
10a+b = 10b+a - 45
Now just crank it out
Please that 10a + b where is it come from in reference to question?
consider the number 47
The digits are 4,7
its value is 10*4 + 7
better review base-10 numbers
a 3-digit number abc has the value 100a+10b+c
and so on
To find the two digits in the given number, let's assume that the tens digit is represented by 'x' and the unit digit is represented by 'y'.
According to the given information, the two-digit number is three times the sum of its digits. This can be expressed as:
10x + y = 3(x + y)
Simplifying this equation, we get:
10x + y = 3x + 3y
Rearranging the terms, we obtain:
10x - 3x = 3y - y
7x = 2y
Now, it is also mentioned that the given number is 45 less than the number formed when the digits are interchanged. This can be expressed as:
10x + y = 10y + x - 45
Rearranging the terms, we have:
10x - x = 10y - y - 45
9x = 9y - 45
Simplifying this equation, we get:
x = y - 5
Now, we have two equations:
7x = 2y
x = y - 5
Substituting the value of x from the second equation into the first equation, we get:
7(y - 5) = 2y
Expanding and simplifying the equation, we have:
7y - 35 = 2y
Moving all the terms to one side, we get:
7y - 2y = 35
Simplifying further, we obtain:
5y = 35
Dividing both sides of the equation by 5, we find:
y = 7
Now, substituting this value of y back into the equation x = y - 5, we get:
x = 7 - 5
x = 2
Therefore, the two digits in the number are 2 and 7.