The sum of the digits of number of two digit number is 12 . If the digits are alter the new number is less by 18 from first number find the number
Answers
Answer:
57
Step-by-step explanation:
If the first digit is represented as a and the second is represented as b:
Then the number n = 10a + b and it’s reverse is 10b + a.
The rules are:
a + b = 12
And
(10b + a) - (10a + b) = 18
This simplified is:
10b + a - 10a - b = 18
9b - 9a = 18
We can factor out 9
9(b - a) = 18 and simplify
b - a = 2
b = 2 + a we can use this to solve for a using:
a + b = 12
a + (2 + a) = 12 Simplify
2a + 2 = 12
2a = 10
a = 5 Now we solve for b
b - a = 2
b - (5) = 2
b = 7
So a = 5 and b = 7 and we can find the number n by using the ewuation we had before:
n = 10a + b
n = 10(5) + 7
n = 50 + 7
n = 57
The answer is 57.
Answer:
Let the tens digit of the required number be x and the units digit be y. Then,
x + y = 12 ---------- (1)
Required Number = (10x+y).
Number obtained on reversing the digits = (10y+x).
Therefore,
(10y + x) - (10x + y) = 18
9y - 9x = 18
y - x = 2 -----------(2)
On adding (1) and (2), we get,
2y = 14
y = 14/2
y = 7
Therefore,
x = 5
Hence, the required number is 57.
Hope it helps