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The sum of 225 consecutive positive integers only has digits 1 or 0. What is the smallest possible value of the sum?

 Jul 28, 2020
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The sum of the sequence must be divisible by 225, or rather multiples of 225 that end in at least 2 zeros. So, will list multiples of 225: 225, 450, 675, 900. Since ALL the digits of the sum of this sequence are only 1's and zeros, then the sum must have a minimum of 9 ones in order to be divisible by 9.Or: 111111111. But we want multiple of 900, so we just add 2 zeros at the end of 9 ones: 11,111, 111,100. This is the sum of your arithmetic sequence.

 

Now, we can use the formula for the sum of an AP to find out the FIRST term:

11,111,111,100 = 225/ 2 * [2*F + 1*(225 - 1)]

                         = 112.5 * [2*F + 224]

                         = 225F + 25,200

225F =11,111,085,900

F = 11,111,085,900 / 225

 

F =49,382,604 and this is your FIRST term of the sequence. If you start with this term and add 1 to each subsequent term for 225 terms until you get the LAST term, or 49,382, 828, and sum them all up, you should get:11,111,111,100. And that is the END of this question! I wonder who thought of this?.

 Jul 28, 2020

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