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BMAD

Summing a summed sequence

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BMAD    62

using the sequence of numbers from 1 to 10,000, sum the digits together and rewrite the numbers as a new sequence.

For example:

......., 345, 346, 347, ....., 5088, 5089,

becomes

........., 12, 13, 14, ......, 21, 22

find a systematic approach or shortcut to adding the numbers from this second sequence without brute force calculations.

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10 answers to this question

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bonanova    76

1 + 2 + 3 + ... + 9,998 + 9,999 + 10,000 =
(1+10,000) + (2+9,999) + (3+9,998) + ... + (4,998+5,003) + (4,999+5,002) + (5,000+5,001) =

5,000 x 10,001 = 50005000.

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BMAD    62

1 + 2 + 3 + ... + 9,998 + 9,999 + 10,000 =

(1+10,000) + (2+9,999) + (3+9,998) + ... + (4,998+5,003) + (4,999+5,002) + (5,000+5,001) =

5,000 x 10,001 = 50005000.

you are adding the actual numbers. Like what Witzar did is what is asked for. 10,000 becomes 1 (since 1+0+0+0+0 = 1) and 9,999 becomes (36 since 9 +9 +9 +9 = 36) change all of the numbers in this manner, then add them up.

but rather then simply doing brute force calculations or using coding, how can we quickly find this solution

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BMAD    62

4*10^(4-1)*(0+1+2+...+9) + 1 = 180001

i like this approach. What is your reasoning?

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witzar    18

4*10^(4-1)*(0+1+2+...+9) + 1 = 180001

i like this approach. What is your reasoning?

All we have to do is to calculate the following sum:

(0+0+0+0) + (0+0+0+1) + (0+0+0+2) + ... + (0+3+4+5) + (0+3+4+6)+ (0+3+4+7) + ... + (5+0+8+8) + (5+0+8+9) + ... + (9+9+9+9)

We obviously have 4*10^4 digits to add, and each digit obviously appears with equal frequency.

Therefore we have 4*10^3 0's, 4*10^3 1's, etc, so the sum equals

4*10^3*(0+1+2+...+9).

Just don't forget to add 1 since we skipped the last term (10000) from original problem.

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DeGe    9

Each digit repeats itself in each position as follows:

in ones position 1000 times

in tens position 100 times

in hundreds position 10 times

in 1000 position 1 times

and 1 appears once in 10 000

Now, 0 + 1 + 2 +3 .... + 9 = 45

So, sum of digits is 45*1000 + 45*100 + 45*10 + 45*1 + 1 = 45*1111 + 1 = 49996

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BMAD    62

Each digit repeats itself in each position as follows:

in ones position 1000 times

in tens position 100 times

in hundreds position 10 times

in 1000 position 1 times

and 1 appears once in 10 000

Now, 0 + 1 + 2 +3 .... + 9 = 45

So, sum of digits is 45*1000 + 45*100 + 45*10 + 45*1 + 1 = 45*1111 + 1 = 49996

i believe that this is the sum of a different sequence

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witzar    18

Each digit repeats itself in each position as follows:

in ones position 1000 times

in tens position 100 times

in hundreds position 10 times

in 1000 position 1 times

and 1 appears once in 10 000

Now, 0 + 1 + 2 +3 .... + 9 = 45

So, sum of digits is 45*1000 + 45*100 + 45*10 + 45*1 + 1 = 45*1111 + 1 = 49996

i believe that this is the sum of a different sequence

The sequence is the same, but there is an error in reasoning.

Each digit repeats itself in each position as follows:

in ones position 1000 times

True.

in tens position 100 times

in hundreds position 10 times

in 1000 position 1 times

3 x False.

and 1 appears once in 10 000

True.

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