SUM OF THE DIGITS OF A TWO DIGIT NUMBER IS 9 AND THE NUMBER ITSELF IS EQUAL TO TWICE THE SUM OF ITS DIGITS. FIND THE NUMBER....(please solve this sum I am you mark as brilliant)
Answers
Given
Sum of digits of a two digit number = 9
Number = Twice the sum of digits.
To find
Original twi digit number
Solution
Let the digit at unit's place be r & digit at ten's place be m.
➥ Original number = r + 10m
According to Question :
➝ r + m = 9
➝ r = 9 - m -(eq.i)
Going to case 2 :
⟼ r + 10m = 2(r + m)
⟼ r + 10m = 2 × 9
⟼ r + 10m = 18
- Putting value from (eq.i)
⟼ 9 - m + 10m = 18
⟼ 9 + 9m = 18
⟼ 9m = 18 - 9
⟼ 9m = 9
⟼ m = 9/9
⟼ m = 1
Putting value in (eq.i) :
➻ r = 9 - 1
➻ r = 8
Now finding the number :
⇰ Number = 8 + 10(1)
⇰ Number = 8 + 10
⇰ Number = 18
Therefore,
Original two digit number = 18 .
Answer:
Let the Ten Digit be b and One Digit be a
⋆ Original Number = (10b + a)
☯ Sum of Digits :
⇾ Sum of Digits = 9
⇾ a + b = 9
⇾ a = 9 – b ⠀— eq. ( I )
☯ According to the Question :
⇢ Number = 2(Sum of Digits)
⇢ (10b + a) = 2(a + b)
- Putting value of Sum of Digits
⇢ 10b + a = 2 × 9
⇢ 10b + a = 18
- Putting value of a from eq. ( I )
⇢ 10b + 9 – b = 18
⇢ 10b – b = 18 – 9
⇢ 9b = 9
- Dividing both term by 9
⇢ b = 1
________________
☢ Putting value of b in eq. ( I ) :
⇢ a = 9 – b
⇢ a = 9 – 1
⇢ a = 8
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☢ Original Number Formed :
⟶ Number = 10b + a
⟶ Number = 10(1) + 8
⟶ Number = 10 + 8
⟶ Number = 18
∴ Hence, Required Number will be 18.