Question

Let a, b, c, d, e be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and b + c + d is square of an integer. The least possible value of the number of digits of c is?
(a) 2
(b) 3
(c) 4
(d) 5

Answer 1

Hint: We know that in an arithmetic series, the numbers are in such sequence that the difference between two consecutive terms is constant throughout the series.

Complete step-by-step answer:
Let a, b, c, d, e be natural numbers in an arithmetic progression. We know that arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between two consecutive terms is constant.
Let us assume that the common difference between them is D and the first term of series is A.
So according to our assumption, the numbers can be written as,
a = A – 2D
b = A – D
c = A
d = A + D
e = A + 2D
We know that it is given in the question that $a+b+c+d+e$ is the cube of an integer. So, let us suppose that the integer is p.
So, we can write that,
$a+b+c+d=e={{p}^{3}}....\left( i \right)$
Also, it is given in the question that b + c + d is the square of an integer. So, let us suppose that the integer is q.
So, we can write that,
$b+c+d={{q}^{2}}....\left( ii \right)$
Now we can substitute the values of A, B, C and D in equation (i) as shown below,
$A-2D+A-D+A+A+D+A+2D={{p}^{3}}$
On simplifying the above equation, we can observe that the variable D will get cancelled out and we will be left with the variable A in the left-hand side of the equation.
$5A={{p}^{3}}....\left( iii \right)$
Now substituting the value of B, C and D in equation (ii), we will get
$A-D+A+A+D={{q}^{2}}$
Again, on simplifying the above equation, we can observe that the variable D will get cancelled out and we will be left with variable A in the left-hand side of the equation.
$3A={{p}^{2}}....\left( iv \right)$
Now we have to divide equation (iv) by equation (iii). We will get,
$\dfrac{5A}{3A}=\dfrac{{{p}^{3}}}{{{q}^{2}}}$
We can observe that variable A is cancelled out and we will be left with
$\dfrac{5}{3}=\dfrac{{{p}^{3}}}{{{q}^{2}}}$
$\dfrac{{{p}^{3}}}{5}=\dfrac{{{q}^{2}}}{3}$
It is said in the question that we have to find the least possible value of the number of digits of c.
So by hit and trial method, the least possibility is
$\begin{aligned} & p=5\times 3 \\\ & q=5\times 3\times 3 \\\ \end{aligned}$
As it will satisfy the equation, we have $p=15\text{ and }q=45$.
Now we know that
$5A={{p}^{3}}$
So putting the value of p = 15 in it, we will get
$\begin{aligned} & 5A={{15}^{3}} \\\ & 5A=3375 \\\ & A=675 \\\ \end{aligned}$
Since we have the value of c = A, we get that c = 675.
It consists of 3 numbers of digits, so the answer to the question is 3. Hence, we get option b as the correct answer.

Note: We can approach this question with a different method. We can take elements of the A.P series as $a,a+d,a+2d...$ instead of $a=A-2D,\text{ }b=A-D,c=A,d=A+D,e=A+2D$. But, this method will increase the number of unknown variables in the equation and make it difficult to solve the question.