987654321 squared. Worked on a 8-digit 4 function inexpensive simple calculator

98765432 1 2 \large 987654321^2

This can not be done directly on a 8-digit 4 function inexpensive simple calculator. But, it can be done indirectly .

Do not use Wolfram/Alpha or Unix's bc or some other extended precision calculator. A Friden mechanical calculator can do it also; but that would be cheating also.

975461027789971041 975461057889971041 975461657789971041 975461057789971041

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3 solutions

Chew-Seong Cheong
Mar 29, 2020

We can rewrite the number 987654321 987654321 as an arithmetic-geometric progression as follows:

987654321 = n = 0 8 1 0 n ( n + 1 ) = 8 × 1 0 10 + 1 9 2 98765432 1 2 = 64 × 1 0 20 + 16 × 1 0 10 + 1 9 4 = 6 , 400 , 000 , 000 , 160 , 000 , 000 , 001 9 4 And using paper and pencil to = 711 , 111 , 111 , 128 , 888 , 888 , 889 9 3 do the long division carefully. = 79 , 012 , 345 , 680 , 987 , 654 , 321 9 2 No calculator needed. = 8 , 779 , 149 , 520 , 109 , 739 , 369 9 = 975 , 461 , 057 , 789 , 971 , 041 \begin{aligned} 987654321 & = \sum_{n=0}^8 10^n(n+1) = \frac {8\times 10^{10}+1}{9^2} \\ \implies 987654321^2 & = \frac {64 \times 10^{20}+16\times 10^{10}+1}{9^4} \\ & = \frac {6,400,000,000,160,000,000,001}{9^4} & \small \red{\text{And using paper and pencil to}} \\ & = \frac {711,111,111,128,888,888,889}{9^3} & \small \red{\text{do the long division carefully.}} \\ & = \frac {79,012,345,680,987,654,321}{9^2} & \small \red{\text{No calculator needed.}} \\ & = \frac {8,779,149,520,109,739,369}9 \\ & = \boxed{975,461,057,789,971,041} \end{aligned}

Please, explain how you got from n = 0 8 1 0 n ( n + 1 ) \sum _{n=0}^8 10^n (n+1) to 8 1 0 1 0 + 1 ) 9 2 \frac{8*10^10 + 1)}{9^2} . I recognize that the result is a correct statement of the value of the summation.

A Former Brilliant Member - 1 year, 2 months ago

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The general formula for the sum of arithmetic-geometric progression is given in the reference as attached in my solution and here for your convenience. You may want to have a look. We can also derive the formula as follows:

n = 0 9 x n = 1 + x + x 2 + + x n Sum of geometric progression x n + 1 1 x 1 = 1 + x + x 2 + + x n Differentiate both sides with respect to x ( n + 1 ) x n ( x 1 ) ( x n + 1 1 ) ( x 1 ) 2 = 1 + 2 x + 3 x 2 + + n x n 1 Sum of arithmetric-geometric progression n x n + 1 ( n + 1 ) x n + 1 ( x 1 ) 2 = 1 + 2 x + 3 x 2 + + n x n 1 Putting x = 10 and n = 9 8 × 1 0 10 + 1 9 2 = 987654321 \begin{aligned} \sum_{n=0}^9 x^n & = 1 + x + x^2 + \cdots + x^n & \small \blue{\text{Sum of geometric progression}} \\ \implies \frac {x^{n+1}-1}{x-1} & = 1 + x + x^2 + \cdots + x^n & \small \blue{\text{Differentiate both sides with respect to }x} \\ \frac {(n+1)x^n(x-1)-(x^{n+1}-1)}{(x-1)^2} & = 1 + 2x + 3x^2 + \cdots + nx^{n-1} & \small \blue{\text{Sum of arithmetric-geometric progression}} \\ \frac {nx^{n+1}-(n+1)x^n+1}{(x-1)^2} & = 1 + 2x + 3x^2 + \cdots + nx^{n-1} & \small \blue{\text{Putting }x=10 \text{ and }n=9} \\ \frac {8\times 10^{10}+1}{9^2} & = 987654321 \end{aligned}

Chew-Seong Cheong - 1 year, 2 months ago

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Thank you. I wondered what I had missed.

A Former Brilliant Member - 1 year, 2 months ago
Elijah L
Mar 28, 2020

This method may be considered cheating.

Through the divisibility test for 9 9 , we know that 987654321 987654321 is divisible by 9 9 . Therefore, we can use the divisibility test for 9 9 on each of the answer responses, and only 975461057789971041 \boxed{975461057789971041} , or the last answer choice, satisfies the divisibility test. (Performing the divisibility test accurately is left as an exercise to the reader.)

Since one out of the four answers is right, it has to be that one.

Touché! That would not have worked if I had changed one of the 9 digits to 0.

A Former Brilliant Member - 1 year, 2 months ago

I calculated using Vedic mathematics.... not too hard.

Nikola Alfredi - 1 year, 2 months ago

Please, explain in detail.

A Former Brilliant Member - 1 year, 2 months ago

Write the number 987654321 987654321 as a base 1000 integer 9, 8765, and 4321.

times 9 8765 4321 9 81 78885 38889 8765 78885 76825225 37873565 4321 38889 37873565 18671041 \begin{array}{r|rrr} \text{times}& 9 & 8765 & 4321 \\ \hline 9 & 81 & 78885 & 38889 \\ 8765 & 78885 & 76825225 & 37873565 \\ 4321 & 38889 & 37873565 & 18671041 \\ \end{array}

Then, extend the numbers with the appropriate number of zeroes and add down the columns:

9 8765 4321 9 810000000000000000 78885000000000000 3888900000000 8765 78885000000000000 7682522500000000 378735650000 4321 3888900000000 378735650000 18671041 + 888888888900000000 86567901235650000 4267654321041 \begin{array}{r|rrr} & 9 & 8765 & 4321 \\ \hline 9 & 810000000000000000 & 78885000000000000 & 3888900000000 \\ 8765 & 78885000000000000 & 7682522500000000 & 378735650000 \\ 4321 & 3888900000000 & 378735650000 & 18671041 \\ \hline + & 888888888900000000 & 86567901235650000 & 4267654321041 \end{array}

Then, add the column sums: 888888888900000000 86567901235650000 4267654321041 975461057789971041 \begin{array}{r} 888888888900000000 \\ 86567901235650000 \\ 4267654321041 \\ \hline 975461057789971041\end{array}

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