9 8 7 6 5 4 3 2 1 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.
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Please, explain how you got from ∑ n = 0 8 1 0 n ( n + 1 ) to 9 2 8 ∗ 1 0 1 0 + 1 ) . I recognize that the result is a correct statement of the value of the summation.
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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 ⟹ x − 1 x n + 1 − 1 ( x − 1 ) 2 ( n + 1 ) x n ( x − 1 ) − ( x n + 1 − 1 ) ( x − 1 ) 2 n x n + 1 − ( n + 1 ) x n + 1 9 2 8 × 1 0 1 0 + 1 = 1 + x + x 2 + ⋯ + x n = 1 + x + x 2 + ⋯ + x n = 1 + 2 x + 3 x 2 + ⋯ + n x n − 1 = 1 + 2 x + 3 x 2 + ⋯ + n x n − 1 = 9 8 7 6 5 4 3 2 1 Sum of geometric progression Differentiate both sides with respect to x Sum of arithmetric-geometric progression Putting x = 1 0 and n = 9
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Thank you. I wondered what I had missed.
This method may be considered cheating.
Through the divisibility test for 9 , we know that 9 8 7 6 5 4 3 2 1 is divisible by 9 . Therefore, we can use the divisibility test for 9 on each of the answer responses, and only 9 7 5 4 6 1 0 5 7 7 8 9 9 7 1 0 4 1 , 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.
I calculated using Vedic mathematics.... not too hard.
Please, explain in detail.
Write the number 9 8 7 6 5 4 3 2 1 as a base 1000 integer 9, 8765, and 4321.
times 9 8 7 6 5 4 3 2 1 9 8 1 7 8 8 8 5 3 8 8 8 9 8 7 6 5 7 8 8 8 5 7 6 8 2 5 2 2 5 3 7 8 7 3 5 6 5 4 3 2 1 3 8 8 8 9 3 7 8 7 3 5 6 5 1 8 6 7 1 0 4 1
Then, extend the numbers with the appropriate number of zeroes and add down the columns:
9 8 7 6 5 4 3 2 1 + 9 8 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 7 8 8 8 5 0 0 0 0 0 0 0 0 0 0 0 0 3 8 8 8 9 0 0 0 0 0 0 0 0 8 8 8 8 8 8 8 8 8 9 0 0 0 0 0 0 0 0 8 7 6 5 7 8 8 8 5 0 0 0 0 0 0 0 0 0 0 0 0 7 6 8 2 5 2 2 5 0 0 0 0 0 0 0 0 3 7 8 7 3 5 6 5 0 0 0 0 8 6 5 6 7 9 0 1 2 3 5 6 5 0 0 0 0 4 3 2 1 3 8 8 8 9 0 0 0 0 0 0 0 0 3 7 8 7 3 5 6 5 0 0 0 0 1 8 6 7 1 0 4 1 4 2 6 7 6 5 4 3 2 1 0 4 1
Then, add the column sums: 8 8 8 8 8 8 8 8 8 9 0 0 0 0 0 0 0 0 8 6 5 6 7 9 0 1 2 3 5 6 5 0 0 0 0 4 2 6 7 6 5 4 3 2 1 0 4 1 9 7 5 4 6 1 0 5 7 7 8 9 9 7 1 0 4 1
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We can rewrite the number 9 8 7 6 5 4 3 2 1 as an arithmetic-geometric progression as follows:
9 8 7 6 5 4 3 2 1 ⟹ 9 8 7 6 5 4 3 2 1 2 = n = 0 ∑ 8 1 0 n ( n + 1 ) = 9 2 8 × 1 0 1 0 + 1 = 9 4 6 4 × 1 0 2 0 + 1 6 × 1 0 1 0 + 1 = 9 4 6 , 4 0 0 , 0 0 0 , 0 0 0 , 1 6 0 , 0 0 0 , 0 0 0 , 0 0 1 = 9 3 7 1 1 , 1 1 1 , 1 1 1 , 1 2 8 , 8 8 8 , 8 8 8 , 8 8 9 = 9 2 7 9 , 0 1 2 , 3 4 5 , 6 8 0 , 9 8 7 , 6 5 4 , 3 2 1 = 9 8 , 7 7 9 , 1 4 9 , 5 2 0 , 1 0 9 , 7 3 9 , 3 6 9 = 9 7 5 , 4 6 1 , 0 5 7 , 7 8 9 , 9 7 1 , 0 4 1 And using paper and pencil to do the long division carefully. No calculator needed.