THE UNIT DIGIT OF A TWO DIGIT NUMBER IS 3 AND SEVEN TIMES THE SUM OF THE DIGIT IS THE NUMBER ITSELF. FIND THE NUMBER
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63 logic is let the first digit be a Then 7(a+3)=10a+3 solving to a=6
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Answer:
Unit of two digit number is 3
Let the tens digit number be x
The Number will be 10x + 3
Sum of two digit number = x + 3
According to the question now :
7(x + 3) = 10x + 3
7x + 21 = 10x + 3
21 - 3 = 10x - 7x
18 = 3x
x = 18/3
x = 6
Therefore, the tens digit number is 6.
Hence,
- The required number = 10x + 3 = 10(6) + 3 = 63.
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