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

Hint:
Here in this question we have to use a simple calculation to find out the original number. The digits of the three digits number are in A.P. (Arithmetic progression). Arithmetic progression (A.P.) is the sequence of numbers such that the difference between the consecutive numbers remains constant. Firstly we have to note down all the conditions given in the question and by simply solving those conditions we will be able to find out the original number.

Complete step by step solution:
Let \[{\rm{(a}} - {\rm{d),a,(a}} + {\rm{d)}}\]be the unit place digit, tens place digit and hundreds place digit respectively.
So, we can write the number as \[{\rm{(a}} + {\rm{d)}} \times 100 + {\rm{a}} \times 10 + {\rm{(a}} - {\rm{d)}} \times 1 = 111{\rm{a + 99d}}\]
Now when the number is reversed we get \[{\rm{(a}} + {\rm{d),a}},{\rm{(a}} - {\rm{d)}}\] as the unit place digit, tens place digit and hundreds place digit respectively.
So, we can write the reversed number as \[{\rm{(a}} - {\rm{d)}} \times 100 + {\rm{a}} \times 10 + {\rm{(a + d)}} \times 1 = 111{\rm{a}} - {\rm{99d}}\]
It is also given that the sum of all the digits is 15, we get
\[ \Rightarrow {\rm{(a}} - {\rm{d)}} + {\rm{a}} + {\rm{(a + d) = 15}}\]
\[ \Rightarrow {\rm{3a = 15}}\]
\[ \Rightarrow {\rm{a = 5}}\]
So, according to the question when the digits are reversed, the number obtained is 594 less than the original number which means
\[111{\rm{a}} - {\rm{99d = }}111{\rm{a + 99d}} - 594\]
Now by solving the above equation, we get
\[ \Rightarrow 99{\rm{d + }}99{\rm{d}} = 594\]
\[ \Rightarrow {\rm{198d}} = 594\]
\[ \Rightarrow {\rm{d}} = 3\]
Therefore, we get the value of both \[{\rm{a = 5}}\]and \[{\rm{d}} = 3\]
So, the original number is\[{\rm{(a}} + {\rm{d)}} \times 100 + {\rm{a}} \times 10 + {\rm{(a}} - {\rm{d)}} \times 1 = 111{\rm{a + 99d}}\] by putting the value of a and d in the expression we will get the original number.
Original number\[{\rm{ = }}111{\rm{a + 99d = (111}} \times {\rm{5) + (99}} \times {\rm{3) = 555 + 297 = 852}}\]

Hence, 852 is the original number.

Note:
Integers are the numbers which can be positive or negative but integers can never be in fractional form or decimal form. Natural numbers are the positive integer numbers.
We always have to keep in mind the basic A.P. and G.P series
A.P. series is \[{\rm{a}},{\rm{a}} + {\rm{d}},{\rm{a}} + 2{\rm{d}},{\rm{a}} + 3{\rm{d,}}............\]
\[{{\rm{n}}^{th}}{\rm{term }} = {\rm{ a + }}\left( {{\rm{n - 1}}} \right){\rm{d}}\]
Where a is the first term and d is the common difference
G.P. series is \[{\rm{a}},{\rm{ar}},{\rm{a}}{{\rm{r}}^2},{\rm{a}}{{\rm{r}}^3},{\rm{a}}{{\rm{r}}^4},.........\]
\[{{\rm{n}}^{{\rm{th}}}}{\rm{term = a}}{{\rm{r}}^{{\rm{n - 1}}}}\]
Where, a is the first term of the G.P. and r is the common ratio.