A number is three times the sum of its digit. What is the number?👇🏾👇 🏾
Let the number be denoted as
ab
, where
a and b are the digits. Since ab
represents a two-digit number, it can be expressed mathematically as:
10a+ba+b10a+b=3(a+b)10a+b=3a+3b3a+3b from both sides:10a+b−3a−3b=07a−2b=0b:7a=2b⟹b=27a
According to the problem, the number is three times the sum of its digits. The sum of its digits is:
Thus, we have the equation:
Expanding and simplifying this equation:
Subtract
This simplifies to:
Solving for
Since
a and b are digits, a must be chosen such that b is an integer between 0 and 9. Therefore, 7a must be divisible by 2, meaning a must be even.a:b=27×2
We test even values of
a=2:
=7⟹Number is 27b=27×4
a=4:
=14⟹Invalid since b is not a digita since they would yield b greater than 9.27=3×(2+7)=3×9=2727Thus, the number that satisfies the given condition is: 27 [ans 27]
No need to test higher even values of
So, we verify the candidate number 27:
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