[jaj2]

Write a function that prints the sum of all the numbers from one to a million whose digits consist only of 2s and 9s. For example, 2, 9, 29, 92922, 222, and 99999 will be in your sum, but not 290, 942, or 4683.

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please post the code you have so far.
hint: use long for sum instead of int

were you able to figure out the sum problem with 1s and 3s?

Edited by bdlt, 28 January 2006 - 02:24 PM.

[bdlt]

please post the code you have so far.
hint: use long for sum instead of int

were you able to figure out the sum problem with 1s and 3s?

Edited by bdlt, 28 January 2006 - 02:24 PM.

[jaj2]

Thanks for the help, I was able to figure it out.

[bdlt]

thanks for letting us know.

FYI -
last last number displayed was 499 993 500 021 - did you get this too?

the final sum(not displayed - add code to display it) should be 500 000 500 000

how do we know this final sum is correct?

for a sum from 1 to n of 1 + 2 + 3 + ... + n = n * (n + 1) / 2

multiply 1000000 * 1000001 then divide by 2 = 500 000 500 000

[destin]

FYI -
last last number displayed was 499 993 500 021 - did you get this too?

the final sum(not displayed - add code to display it) should be 500 000 500 000

how do we know this final sum is correct?

for a sum from 1 to n of 1 + 2 + 3 + ... + n = n * (n + 1) / 2

multiply 1000000 * 1000001 then divide by 2 = 500 000 500 000

What? If we're looping one million times then how can the last number be greater than one million (499,993,500,021)? In the OP, it was stated that you can only use 9's and 2's. The last number will be 999,999. 500,000,500,000 is the sum of all numbers from 1 to 1000000 (inclusive)... this doesn't relate to the OP either. The sum I got was 41,169,513. Please verify this... I'm lost.

[bdlt]

instead of "last last number displayed" it would be more clear as "last sum displayed".

my error - "whose digits consist only of 2s and 9s". didn't 'read' 2s AND 9s(and misread examples 2, 9, 29,...). so I'll post a new 2s and 9s ONLY number.

"The sum I got was 41,169,513. " - think about it. the last 10 numbers summed will almost reach 10 million, so 41 million is way too low.

edit - it looks like there may be NO sums whose digits consist only of 2s and 9s. did you get this too?

Edited by bdlt, 04 February 2006 - 10:06 PM.

[destin]

"The sum I got was 41,169,513. " - think about it. the last 10 numbers summed will almost reach 10 million, so 41 million is way too low.

So what'd you get?

[bdlt]

this should be the sum: 500 000 500 000

and it looks like there are NO sums whose digits consist only of 2s and 9s. did you get this too?

[destin]

this should be the sum: 500 000 500 000

How do you figure? As I said before, this is the sum of all numbers from 1 to 1000000.

and it looks like there are NO sums whose digits consist only of 2s and 9s. did you get this too?

What do you mean?

[bdlt]

for a sum from 1 to n of 1 + 2 + 3 + ... + n = n * (n + 1) / 2

multiply 1000000 * 1000001 then divide by 2 = 500 000 500 000

and it looks like there are NO sums whose digits consist only of 2s and 9s. did you get this too?

What do you mean?

there is not any sum that consists of only 2s and 9s.

1+2 = 3
1+2+3 = 6
1+2+3+4 = 10

1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28

and so on

as a test - try sums whose digits are only 0,1,2, and 9(I get 48)

what sums do you see that are only 2s and 9s?

[destin]

I think we are interpreting the question differently. I summed up numbers 2, 9, 22, 29, 92, 99, etc. Reread the OP... I see how it could go both ways, but the last sentence states examples that must be in your final sum.

[bdlt]

got it - the sum is only of numbers with 2s and 9s. we don't care what the sum of ALL numbers may be.

0 points for bdlt on this one.

edit - 41,169,513 looks correct

Edited by bdlt, 04 February 2006 - 10:56 PM.