Let be a positive integer with all digits equal to such that is divisible by . Find the last six digits of .

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- October 16th 2008, 01:24 PMalexmahoneFind the last six digits of...
Let be a positive integer with all digits equal to such that is divisible by . Find the last six digits of .

- October 17th 2008, 03:00 PMAlexgadgetman
55555555555555

And I dont even know how you could use a formula... ah well.. trial and error does work sometimes.

Last six digits are 173518 as total resulting number is

27736173518

Bearing in mind you needed 6 digits at least, so number would need 9(?) 5's at least.

I just used trial and error.... (Worried) - October 17th 2008, 07:13 PMTing
Hello

Here is one way, I think.

If you are familiar with Napier's Bones method of multiplication then you can work 'backwards' to find the digits.

The last digit of must be 5, since that is the only digit, (0 to 9), that will give you a number ending in 5 when multiplied by 3. The next digit must be 8, the only digit that will give you a number ending in 4, when multiplied by 3.

Please see the attachment below.

Can you see how to get the rest of the digits you need?

Hope that helps.