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  • Directions for question : In this fictitious tale, the two legendary mathematicians and astronomers Aryabhatta and Varahamihira met at a d

Directions for question :

In this fictitious tale, the two legendary mathematicians and astronomers Aryabhatta and Varahamihira met at a dinner party hosted by King Chandragupta Vikramaditya at his palace in Pataliputra. 

Aryabhatta, the senior and more moody among the two, being slightly inebriated with the excellent ‘somras’ served by the king, and in a very good mood, suddenly said to Varahamihira -- ‘If distinct alphabets of the english language stand for distinct digits, then SATURN plus URANUS will be equal to JUPITER’!! 

Though initially puzzled by this sudden statement from Aryabhatta, Varahamihira quickly regained composure and solved the challenge thrown by Aryabhatta.

Question :

What would the ‘SUN’ be represented by as per the thought process of Aryabhatta ?

Option: 1

957


Option: 2

864


Option: 3

764


Option: 4

569


Answers (1)

best_answer

As per the statement of Aryabhatta,

    \begin{array}{rccccc} S & A & T & U & R & N \\ +U & R & A & N & U & S \\ \hline J U & P & I & T & E & R \end{array}

A seven digit number added to another seven digit number has given an eight digit number. Hence J must be equal to 1, as the maximum value of S+U can be 18.

S+U can never be equal to a two digit number(</= 18)  whose units digit is U, unless a carrying of 1 has been added to it, and unless S is equal to 9.

Hence,

    \begin{array}{rccccc} 9 & A & T & U & R & N \\ +U & R & A & N & U & 9\\ \hline 1 U & P & I & T & E & R \end{array}

N+9 would be equal to 9 (that is S) if N is equal to 0. Since it is not so, hence N is not equal to 0. If N is any other digit, then R+1 must be equal to N, and N+9 must be a two digit number (</= 18)

So R+U+1 is equal to a two digit number (</= 18) with units digit E, because In the next column, U+N, that is U+R+1, must have a carrying of 1 added to it to be equal to a different two digit number (</= 18) with units digit T Also we can conclude T is equal to E+1.
Hence,     

  \begin{array}{r} 9 A(E+1) \cup(N-1) N \\ +\cup(N-1) A N \cup \quad 9 \\ \hline 1 \cup P \quad I(E+1) E(N-1) \end{array}
\begin{aligned} & 1) \text{ Let } \mathrm{N}\text{ be equal to } 8 . \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & U & 7 & 8 \\ +U & 7 & A & 8 & U & 9 \\ \hline 1 U & P & I(E+1) & E & 7 \end{array} \end{aligned}

\begin{aligned} & \text { a) Let } U \text { be equal to } 0 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 0 & 7 & 8 \\ +0 & 7 & A & 8 & 0 & 9 \\ \hline 1 \ 0 & P & I(E+1) & E & 7 \end{array} \end{aligned}

But then E becomes equal to 8, which is not possible, as N is equal to 8.

\begin{aligned} & \text { b) Let } U \text { be equal to } 2 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 2 & 7 & 8 \\ +2 & R & A & 8 & 2 & 9 \\ \hline 1 \ 2 & P & I(E+1) & E & 7 \end{array} \end{aligned}

So E becomes equal to 0, and T equal to 1, which is not possible, as J is equal to 1

\begin{aligned} & \text { c) Let } U \text { be equal to } 3 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 3 & 7 & 8 \\ +3 & 7 & A & 8 & 3 & 9 \\ \hline 1 \ 3 & P & I(E+1) & E & 7 \end{array} \end{aligned}

But then E becomes equal to 1, which is not possible, as J is equal to 1
\begin{aligned} & \text { d) Let } U \text { be equal to } 4 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 4 & 7 & 8 \\ +4 & 7 & A & 8 & 4 & 9 \\ \hline 1 \ 4 & P & I(E+1) & E & 7 \end{array} \end{aligned}

But then E becomes equal to 2, and T is equal to 3. Hence, 

\begin{aligned} & \begin{array}{rrrrrr} 9 & A&3 & 4 & 7 & 8 \\ +4 & 7 & A & 8 & 4 & 9 \\ \hline 1 \ 4 & P & I &3 & 2& 7 \end{array} \end{aligned}

But the remaining digits of 5, 6 and 0 can never substitute A, P or I in any order and produce the desired result as given.

\begin{aligned} & \text { e) Let } U \text { be equal to } 5 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 5 & 7 & 8 \\ +5 & 7 & A & 8 & 5 & 9 \\ \hline 1 \ 5 & P & I(E+1) & E & 7 \end{array} \end{aligned}

But then E becomes equal to 3, and T equal to 4. Hence,    

\begin{array}{rrrrrr} 9 & A&4 & 5 & 7 & 8 \\ +5 & 7 & A & 8 & 5 & 9 \\ \hline 1 \ 5 & P & I &4 & 3 & 7 \end{array}

But the remaining digits of 2, 6 and 0 can never substitute A, P or I in any order and produce the desired result as given.

\begin{aligned} & \text { f) Let } U \text { be equal to } 6 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 6 & 7 & 8 \\ +6 & 7 & A & 8 & 6 & 9 \\ \hline 1 \ 6 & P & I(E+1) & E & 7 \end{array} \end{aligned}

But then E becomes equal to 4, and T equal to 5. Hence, 

\begin{array}{rrrrrr} 9 & A&5 & 6 & 7 & 8 \\ +6 & 7 & A & 8 & 6 & 9 \\ \hline 1 \ 6 & P & I &5 & 4 & 7 \end{array}

But the remaining digits of 2, 3 and 0 can never substitute A, P or I in any order and produce the desired result as given.

g) U can have no other value, as 7, 8 and 9 are already taken.

\begin{aligned} & \text { 2) Let } U \text { be equal to } 7 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & U & 6 & 7 \\ +U & 6 & A & 7 & U & 9 \\ \hline 1 \ U & P & I(E+1) & E & 6 \end{array} \end{aligned}\begin{aligned} & \text { a) Let } U \text { be equal to } 0 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 0 & 6 & 7 \\ +0 & 6 & A & 7 & 0 & 9 \\ \hline 1 \ 0 & P & I(E+1) & E & 6 \end{array} \end{aligned}

 

But then E becomes equal to 7, which is not possible, as N is equal to 7.\begin{aligned} & \text { b) Let } U \text { be equal to } 2 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 2 & 6 & 7 \\ +2 & 6 & A & 7 & 2 & 9 \\ \hline 1 \ 2 & P & \ I&(E+1) & E & 6 \end{array} \end{aligned}

So E becomes equal to 9, which is not possible, as S is equal to 9

\begin{aligned} & \text { b) Let } U \text { be equal to } 3 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 3 & 6 & 7 \\ +3 & 6 & A & 7 & 3 & 9 \\ \hline 1 \ 3 & P & \ I&(E+1) & E & 6 \end{array} \end{aligned}

But then E becomes equal to 0, and T is equal to 1, which is not possible, as J is equal to 1\begin{aligned} & \text { c) Let } U \text { be equal to } 4 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 4 & 6 & 7 \\ +4 & 6 & A & 7 & 4 & 9 \\ \hline 1 \ 4 & P & \ I&(E+1) & E & 6 \end{array} \end{aligned}

But then E becomes equal to 1, which is not possible, as J is equal to 1\begin{aligned} & \text { d) Let } U \text { be equal to } 5 \text {. } \\ \\ & \begin{array}{rrrrrr} 9 & A&(E+1) & 5 & 6 & 7 \\ +5 & 6 & A & 7 & 5 & 9 \\ \hline 1 \ 5 & P & \ I&(E+1) & E & 6 \end{array} \end{aligned}But then E becomes equal to 2, and T equal to 3. Hence,  \begin{array}{rrrrrr} 9 & A&3 & 5 & 6 & 7 \\ +5 & 6 & A & 7 & 5 & 9 \\ \hline 1 \ 5 & P & \ I&3 & 2 & 6 \end{array}

The remaining digits are 4, 8 and 0. If A is equal to 4, we get

\begin{array}{rrrrrr} 9 & 4&3 & 5 & 6 & 7 \\ +5 & 6 & 4 & 7 & 5 & 9 \\ \hline 1 \ 5 & 0 & \ 8&3 & 2 & 6 \end{array}

The desired result is reached.

Hence, Varahamihira found out that the ten letters of the english alphabets Aryabhatta had used in his statement of SATURN + URANUS = JUPITER were in place of the following digits : 

A = 4,

E = 2,

I = 8,

J = 1,

N = 7,

P = 0,

R = 6,

S = 9,

T = 3, and

U = 5

Hence, as per the thought process of Aryabhatta, the ‘SUN’ be represented by 957


 

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manish

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