# 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

Last updated date: 25th Mar 2023

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Answer

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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{align}

& p=5\times 3 \\

& q=5\times 3\times 3 \\

\end{align}$

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{align}

& 5A={{15}^{3}} \\

& 5A=3375 \\

& A=675 \\

\end{align}$

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.

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{align}

& p=5\times 3 \\

& q=5\times 3\times 3 \\

\end{align}$

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{align}

& 5A={{15}^{3}} \\

& 5A=3375 \\

& A=675 \\

\end{align}$

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.

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