Question

The digits of a positive number of 3 digits area in A.P. and their sum is 15. The number obtained by reversing the digit is 594 less. Find the number!! fast.. Tomorrow exam!

Answer 1

Hello!

$\underline{\bf{Let}}$ :-

• The digit at the tens place be $\red{\bf{a}}$.
• The digit at the hundredth place be $\red{\bf{(a + d)}}$
• The digit at the ones place be $\red{\bf{(a - d)}}$

Then,

$\underline{\bf{Sum\: of \:the \:digits}}$ :

(a + d) + a + (a – d) = 15 

⇒ 3a = 15 ⇒ a = $\dfrac{15}{3}$ ⇒ $\boxed{\bf{a = 5}}$

Now :

⇒ (555 + 99d) – (555 - 99d) = 594

⇒ 198d = 594 ⇒ d = $\dfrac{594}{198}$ ⇒ $\boxed{\bf{d = 3}}$

Therefore,

• a = 5
• (a + d) = (5 + 3) = 8
• (a – d) = (5 – 3) = 2

Hence,
The number formed = $100(8) + 10(5) + 1(2)$ = $\boxed{\blue{\bf{852}}}$

Cheers!

Answer 2

Answer:

Let

:-

• The digit at the tens place be \red{\bf{a}}a .

• The digit at the hundredth place be \red{\bf{(a + d)}}(a+d)

• The digit at the ones place be \red{\bf{(a - d)}}(a−d)

Then,

\underline{\bf{Sum\: of \:the \:digits}}

Sumofthedigits

:

(a + d) + a + (a – d) = 15

⇒ 3a = 15 ⇒ a = \dfrac{15}{3}

3

15

⇒ \boxed{\bf{a = 5}}

a=5

Now :

⇒ (555 + 99d) – (555 - 99d) = 594

⇒ 198d = 594 ⇒ d = \dfrac{594}{198}

198

594

⇒ \boxed{\bf{d = 3}}

d=3

Therefore,

• a = 5

• (a + d) = (5 + 3) = 8

• (a – d) = (5 – 3) = 2

Hence,

The number formed = 100(8) + 10(5) + 1(2)100(8)+10(5)+1(2) = \boxed{\blue{\bf{852}}}

852

Step-by-step explanation:

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