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!!

plssss!

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

let the number be ,a+d,a,a-d

a-d+a+a+d=15

a=5 ...........................i

original number=

(a-d)100+10a+a+d=111a+99d

reversed number

100(a+d)+10a+a-d=111a-99d

atq

111a-99d + 594=111a+99d

=189d=594

d=3 ................ii

from i and ii

111a+99d= 111(5) +99(3)

the number is 852