Question

The digits of a three digit number N are in AP. If the sum of the 15 and the number obtained by reversing the digits of the number 594 less than the original number , then 1000/N -252 is equal to

Answer 1

Correct Question

The digits of a three digit number are in AP and their sum is 15. And the number obtained by reversing the digits of the number 594 less than the original number. Find the number.

Solution

Let the there digit be (a - d), a, (a + d)

Sum of the assumed three digits number is 15.

⇒ (a - d) + a + (a + d) = 15

⇒ a - d + a + a + d = 15

⇒ 3a = 15

Divide by 3 on both sides

⇒ 3a/3 = 15/3

⇒ a = 5

Also, the number obtained by reversing the digits of the number 594 less than the original number.

If we assumed the numbers be x and y (x as tens and y as ones) then the number will be 10x + y and it's reverse will be 10y + x.

But here, we assumed three numbers i.e. (a-d), (a) and (a+d). So, the number will be 100(a - d) + 10(a) + (a + d). And it's reverse becomes 100(a + d) + 10(a) + (a - d).

According to question,

⇒ 100(a - d) + 10(a) + (a + d) - 594 = 100(a + d) + 10(a) + (a - d)

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

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

⇒ - 200d + 2d = 594

⇒ 2(-100d + d) = 594

⇒ - 99d = 297

⇒ d = -3

From the above calculations we have, a = 5 and d = -3

So,

A.P. = (a - d), a, (a + d)

⇒ [5 - (-3)], 5, (5 - 3)

⇒ (5 + 3), 5, (2)

⇒ 8, 5, 2

Therefore,

Original number = 100(a - d) + 10(a) + (a + d) = 800 + 50 + 2 = 852

Answer 2

Given :-

• Three digit Numbers are in AP .
• Sum of Digits = 15.
• Difference b/w, original Number and Reversing obtained Number = 594.

To Find :-

• Original Number ?

Solution :-

Lets Assume That, Three Digits of The Original numbers are (a - d), a & (a + d).

where ,

→ Hundred Place = (a - d).

→ Ten Place = a

→ unit place = (a + d).

Now, sum of 3 digits is given 15.

So,

→ (a - d) + a + (a + d) = 15

→ 3a = 15

→ a = 5 ------------------- Equation (1).

______________________

Now, when Original Number is Reversed , we have ,

→ Hundred Place = (a + d).

→ Ten Place = a

→ unit place = (a - d).

Now, we have Difference b/w, original Number and Reversing obtained Number is 594.

So,

→ Original Number - Reversing Number = 594

Or,

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

→ [ 100a - 100d + 10a + a + d] - [ 100a + 100d + 10a + a - d] = 594

→ [ 111a - 99d ] - [ 111a + 99d ] = 594

→ 111a - 111a - 99d - 99d = 594

→ - 198d = 594

→ d = (-3) ------------------- Equation (2).

_____________________

Putting Values of a & d from Both Equations we get :-

Original number :-

→ Hundred Place = (a - d) = (5 - (-3)) = 5 + 3 = 8

→ Ten Place = a = 5

→ unit place = (a + d) = (5 + (-3)) = 5 - 3 = 2 .

Hence, The original Three Digit Number is 852.