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A persistent divisibility (Posted on 2008-07-03) Difficulty: 2 of 5
A number less than one million is known to be such that diminishing it by 3 makes it divisible by 7.

From the number diminished by 3, we subtract the seventh part. We then obtain a number that also becomes divisible by 7 after 3 is subtracted from it.

From this number we derive another in the same way, namely by subtracting a seventh part from the number diminished by 3.

This time, too, a number results that is divisible by 7 after subtracting 3.

This occurs four more times, so that, in all, it happens seven times over, that a number is divisible by 7 after subtracting 3.

Find this number.

See The Solution Submitted by pcbouhid    
Rating: 2.5000 (2 votes)

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Solution Solution | Comment 1 of 2

Let the number at the end of the ith operation be A(i), for i = 1, 2, ..., 7. Let the number originally before the first operation be denoted by A(0).

Then, by the problem:

A(1) = (A(0) - 3) - (A(0) - 3)/7 = 6/7*(A(0) - 3)

A(2) = (A(1) - 3) - (A(1) - 3)/7 = 6/7*(A(1) - 3),and:

A(i+1) = 6/7*(A(i) - 3), for i = 0, 1, ..., 6

Or, A(i) = (7/6)*A(i+1) + 3 .....(#)

Let A(7) = x, then in terms of (#), we obtain:

A(6) = (7/6)*x + 3

A(5) = (7/6)((7/6)*x + 3) + 3
= (7/6)^2*x + 3*(7/6) + 3

Proceeding in this manner we observe that:

A(1) = (7/6)^6*x + 3*(7/6)^5 +......+ 3, and:

A(0) = (7/6)^7*x + 3*( (7/6)^6 +......+ 1)
= (7/6)^7*x + 3*6*((7/6)^7 - 1)
= (7/6)^7*(x + 18) - 18

or, (A(0) + 18)/(7^7) = (x+18)/(6^7) = v (say), where v is a positive integer.

Thus, A(0) = 7^7*v - 18

Now, by the given conditions:

A(0) < 10^6

or, v < (10^6 + 18)/(7^7) = 1.21429 (correct to 5 places)

The only positive integer satisfying the above inequality occurs at v = 1, giving:

A(0) = 7^7 - 18 = 823,525

Consequently, the required number is 823,525.

Edited on July 3, 2008, 12:49 pm
  Posted by K Sengupta on 2008-07-03 11:55:08

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