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 A Sevens And Zeroes Problem (Posted on 2006-01-26)
Determine the smallest number constituted entirely by sevens and zeroes which is divisible by 61376.

How about for 107072?

 See The Solution Submitted by K Sengupta Rating: 3.5000 (2 votes)

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 And the second part. (Spoiler) | Comment 3 of 6 |

Edited for clarity:

I ran out of fingers for this one.

7,777,777,000,000 = 107072 * (5^6) * 4649

Ady gave the process away. Factoring 107072: 2^6 * 7 * 239.

The final solution with 7's and 0's will by divisible by 7.  In order to reduce the size of the problem, remove the 7 from the initial calculations and look for a number with all 1's and 0's.

The first number of the answer must be a 1.  Solutions will fall between 1.00000 * 10^x and 1.111111(repeating)*10^x.

The factors of 2 can be removed since they will not create an answer that is all 1's or 0's without an equal number of 5's

This gives as a starting point numbers bounded between 4.1841 *10^y and 4.649*10^y (1.000*10^x/239 and 1.1111*10^x/239).

One additional constraint is that the answer will have a 9 in the units place. 9 is the only number that multipied by 9 will result with a 1 in the units place. Multiplication by 0 will result in a 0 in the units place but will not be the lowest value.

Set y = 1 then acceptable values are between 41 and 46. None end in 9

Set y = 2. Acceptable values between 418 and 464. Choosing those ending in 9 . . . gives 419, 429, 439, 449, 459. No solution.

Set y = 3. Values between 4184 and 4649. Trial and error (Excel) gives 4649 as the first number that multiplied by 239 produces all 1's and 0's (1,111,111)

Lumping all the factors together: 7 * 239 * 4649 * (2^6) * (5^6)

= 7 * 1111111 * 10^6 = 7777777000000

Edited on January 27, 2006, 8:50 am
 Posted by Leming on 2006-01-26 16:38:45

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