In a two-digit number ,the digit at the units place is douale the digit in the tens place. The number exceeds the sum of digits by18.find the number.
n = 2 didit number
a = tens
b = ones
So:
n = 10 a + b
The number exceeds the sum of digits by18 mean:
n = a + b + 18
Now equalize exspressions for n.
n = n
10 a + b = a + b + 18
Subtract b to both sides
10 a + b - b = a + b + 18 - b
10 a = a + 18
Subtract a to both sides
10 a - a = a + 18 - a
9 a = 18
Divide both sides by 9
a = 18 / 9
a = 2
The digit at the units place is douale the digit in the tens place mean:
a = 2 b
this mean:
2 = 2 b
Divide both sides by 2
1 = b
b = 1
Your two-digit numer :
n = 21
Proof:
The number exceeds the sum of digits by18.
21 = 2 + 1 + 18
21 = 3 + 18
21
To find the number, we need to analyze the given information step by step.
Let's start with breaking down the problem into smaller parts:
1. The digit in the units place is double the digit in the tens place: Let's assume the digit in the tens place is "x." According to the problem, the digit in the units place would be "2x" since it is double the digit in the tens place.
2. The number exceeds the sum of its digits by 18: The sum of the digits of a two-digit number can be found by adding the tens place digit and the units place digit together. So, the sum of the digits would be "x + 2x".
According to the problem, the number exceeds this sum by 18. In other words, the number can be expressed as "(x + 2x) + 18".
Now we can combine these equations to solve for "x":
x + 2x = (x + 2x) + 18
Simplifying the equation:
3x = 3x + 18
Bringing the like terms to one side:
3x - 3x = 18
0 = 18
Oops! It seems like there's no valid solution for "x" in this case. Something might have gone wrong with the initial conditions or calculations. Please double-check the given information to ensure its accuracy.