Question

The digits of a positive integer, having three digits are in A. P and their sum is 15.The number obtained by reversing the digits is 594 less than the original number. Find the number ​

Answer 1

AnswEr:-

Three digits number = 852

$\rule{200}{2}$

Let the three digits of the number be a, a + d & a + 2d as they are in A.P

∴ Number = a + 10(a + d) + 100(a + 2d)

Case 1:-

⇒ a + (a + d) + (a + 2d) = 15

⇒ a + a + d + a + 2d = 15

⇒ 3a + 3d = 15

⇒ 3(a + d) = 15

⇒ a + d = 15/3

⇒ a + d = 5 [Eq.1]

Case 2:-

Number obtained by reversing digits:-

∴ Number = 100a + 10(a + d) + (a + 2d)

According to question:-

⇒ [a + 10(a + d) + 100(a + 2d)] - [100a + 10(a + d) + (a + 2d)] = 594

⇒ a + 10a + 10d + 100a + 200d - 100a - 10a - 10d - a - 2d = 594

⇒ 111a + 210d - 111a - 12d = 594

⇒ 198d = 594

⇒ d = 594/198

⇒ d = 3

Put this value in (Eq.1)

⇒ a + 3 = 5

⇒ a = 5 - 3

⇒ a = 2

$\rule{150}{1}$

⋆ THREE DIGITS OF THE NUMBER:-

↠ First digit = a = 2

↠ Second digit = a + d = 2 + 3 = 5

↠Third digit = a + 2d = 2 + 2(3)= 2 + 6 = 8

$\rule{150}{2}$

⋆ THREE DIGITS NUMBER :-

⇒ Number = 2 + 10(2 + 3) + 100{2 + 2(3)}

⇒ Number = 2 + 10(5) + 100(8)

⇒ Number = 2 + 50 + 800

⇒ Number = 852

Therefore,

Three digits number = 852 .

Answer 2

Given :

• The digits of a positive integer, having three digits are in A. P.
• The sum of the three digits is 15.
• The number obtained on reversing the digit is 594 less than the original number.

To Find :

• The original number.

Solution :

Let the digit at the hundredth place be (a+d)

Let the digit at the tens place be a

Let the digit at the units place be (a-d)

Original Number = $\bold{100(a+d)+10a+a-d}$

Case 1 :

Sum of the digits of hundredth place, tens place and units place is 15.

Equation :

$\sf{(a+d) +(a)+(a-d)=15}$

$\sf{a+d+a+a-d=15}$

$\sf{3a=15}$

$\sf{a=\dfrac{15}{3}}$

$\sf{a=5\:\:\:\:(1)}$

Case 2 :

When the digits are reversed of the original number, the new number formed is 594 less than the original number.

Reversed Number = $\bold{100(a-d)+10a+a+d}$

Equation :

$\sf{100(a-d) +10a+a+d=100(a+d)+10a+(a-d) -594}$

$\sf{100(5-d)+10(5)+5+d=100(5+d)+10(5)+(5-d)-594}$

$\bold{\big[From\:equation\:(1),\:a\:=\:5\big]}$

$\sf{500-100d+50+5+d=500+100d+50+5-d-594}$

$\sf{500+50+5-100d+d=500+50+5-d-594}$

$\sf{555-100d+d=555+100d-d-594}$

$\sf{-100d+d=100d-d-594}$

$\sf{-99d=99d-594}$

$\sf{-99d-99d=-594}$

$\sf{-198d=-594}$

$\sf{d=\dfrac{-594}{-198}}$

$\sf{d=\dfrac{594}{198}}$

$\sf{d=3\:\:\:(2)}$

Number :

$\large{\boxed{\bold{Hundreth\:digit\:=\:100(a+d)=100(5+3)=100(8)=800}}}$

$\large{\boxed{\bold{Ten's\:digit\:=\:10a\:=\:10(5)=50}}}$

$\large{\boxed{\bold{Unit's\:digit\:=\:(a-d)\:=\:5-3=2}}}$

$\large{\boxed{\bold{\purple{Original\:Three\:digit\:Number\:\:=100(a+d)+10a+a-d=852}}}}$