What two digit number is 3 more than 4 times the sum of its digits if 18 is added to the number its digits are reversed find the number?

A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number its digits are reversed. Find the number.

Answer

Hint: Let the ones digit of the original number be x and the tens digit of the original number be y. So, the original number is 10y+x. So, use the statement that the two digit number is 3 more than 4 times the sum of its digits to form the first equation. When the digits are reversed the number we get is 10x+y which is equal to 18 added to the original number which will give the second equation. Solve the equations to get the answer.

Complete step-by-step answer:
Let the ones digit of the original number be x and the tens digit of the original number be y. So, the original number is 10y+x.
Now, it is given that the two digit number is 3 more than 4 times the sum of its digits, so if we represent this mathematically, we get
 $ 10y+x=3+4\left( x+y \right) $
 $ \Rightarrow 10y+x=3+4x+4y $
 $ \Rightarrow 6y-3x=3.............(i) $
It is also given that when the digits are reversed the number we get is 10x+y which is equal to 18 added to the original number. So, if we represent this as an equation, we get
 $ 10y+x+18=10x+y $
 $ 9x-9y=18...........(ii) $
Now, we will multiply equation (i) by 3 and add it with equation (ii). On doing so, we get
 $ 9x-9y-3\left( 6y-3x \right)=18-3\times 3 $
\[\Rightarrow -9y+18y=18+9\]
\[\Rightarrow y=\dfrac{27}{9}=3\]
If we substitute y in equation (i), we get
 $ 6y-3x=3 $
 $ \Rightarrow 6\times 3-3x=3 $
 $ \Rightarrow 18-3=3x $
 $ \Rightarrow x=\dfrac{15}{3}=5 $
Hence, the original number is 10y+x=30+5=35.

Note: Don’t get confused and consider the original to be yx, where y is a digit and x is another digit, because this will mean that the number is equal to the product of y and x while the number is actually equal to (10y+x). Also, make sure that you are not confused about which variable is the unit digit of the original number and which is the unit digit of the reversed number, as it might be very confusing.

let us consider the unit place digit as x and tens place digit as y.

The equations become 10y + x……..equation (1)

From the question, a two-digit number is 3 more than 4 times the sum of its digits

∴from the above condition, 4(y + x) + 3……… equation (2)

Combining equation 1 and 2

4(y + x) + 3 = 10y + x

4y + 4x + 3 = 10y + x

4x – x + 4y – 10y = -3

3x – 6y = -3

3(x – 2y) = -3

X -2y = -1 …………..equation (3)

From the second condition, If 18 is added to the number, its digits are reversed

∴the reversed number is 10x + y……..equation (4)

∴by the given condition

(10y + x) + 18 = 10x + y

10y – y =10x –x -18

9y – 9x = -18

9(y - x) = -18

Y – X = -2 ……….equation (5)

Solving equation 3 and 5 simultaneously we get,

Y=3 and x = 5

∴the required number is (10y + x) = (10(3) + 5) = 30 + 5 = 35

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
10x + y = 4(x + y) + 3
⇒10x + y = 4x + 4y + 3
⇒ 6x – 3y = 3
⇒ 2x –y = 1                             ……….(i)
Again, we have:
10x + y + 18 = 10y + x
⇒9x – 9y = -18
⇒x – y = -2                              ……..(ii)
On subtracting (ii) from (i), we get:
x = 3
On substituting x = 3 in (i) we get
2 × 3 –y = 1
⇒ y = 6 – 1 = 5
Required number = (10x + y) = 10 × 3 + 5 = 30 + 5 = 35
Hence, the required number is 35.

A two digit number is 3 more than 4 times the sum of its digits.If 18 is added to the number,its digits are reversed.Find the number.Answer:35

Solution

Let the digits be x and y. Thus, number=10x+y Now, (10x+y) - 4(x+y)= 3.....(i) and (10x+y)+ 18=10y+x.....(ii) Solving (i) 10x+y-4x-4y=3 or,6x-3y=3 or,2x-y=1....(1) Solving (ii) 10x+y-10y-x=-18 or, 9y-9x=18 or,y-x=2...(2) (1)+(2) 2x-y+y-x=1+2 or,x=3 Thus, y=x+2 (from (2)) y=5 Therefore the number= 10x+y=10x3+5=35

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