Let M and N be two 9-digit positive integers with the property that?

Question

Let M and N be two 9-digit positive integers with the
property that if any one digit of M is replaced by the digit
of N in the corresponding place (e.g., the `tens' digit of M
replaced by the `tens' digit of N) then the resulting integer is
a multiple of 7.
Prove that any number obtained by replacing a digit of N by
the corresponding digit of M is also a multiple of 7.
Find an integer d > 9 such that the above result concerning
divisibility by 7 remains true when M and N are two d-digit
positive integers.

Answer 1

Actually pretty simple when you get into it, but a nice exercise.

Let k be the remainder when M is divided by 7. Let the j'th digits of M and N be denoted by m_j and n_j respectively. The property will be true if and only if for all j we have
(n_j - m_j) . 10^(9-j) ≡ -k (mod 7).
Then N - M = Σ(n_j - m_j) . 10^(9-j)
≡ 9(-k) (mod 7) ≡ -2k (mod 7), so N ≡ -k (mod 7).
Since for all j we have
(m_j - n_j) . 10^(9-j) ≡ k (mod 7), it follows that replacing a digit of N by the corresponding digit of M also yields a multiple of 7.

It follows from the above analysis that any number of digits which is congruent to 2 (mod 7) will work. So possible values of d > 9 are 16, 23, 30, etc.

Answer 2

a million. This follows from the completeness axiom that announces each and every set of exact numbers with an top sure has a least top sure. you may proceed by ability of contradiction: think there is not any n?N such that nr > s, then nr ? s for all n?N so the set S = {nr | n?N} is bounded above by ability of s. subsequently by ability of the completeness axiom, S has a least top sure M. subsequently M-r can't be an top sure for S and so there exists n? ? N such that M-r < n?r ? M. From this it follows that M < n?r + r, it extremely is M < (n?+a million)r. yet (n?+a million)r ? S, so M isn't an top sure of S, contradicting that M is least top sure of S. This contradiction got here from the theory that the Archimedean property replaced into fake, and so proves that the Archimedean property is actual. 2. For a rational between a and b we could desire to locate advantageous integers m and n such that a < m/n < b, equivalently such that an < m < bn. on account that b-a>0, enable r=b-a and s=a million on your assertion of the Archimedean property. It then says there exists a very good integer n such that n(b-a) >a million, equivalently nb - na > a million, and so there could desire to be some integer m between na and nb. to locate an irrational between a and b, enable m/n be a rational between a and b and positioned r = (b-m/n)?2 and s = a million on your assertion of the Archimedean property. Then by ability of the Archimedean property, there exists a very good integer ok such that kr > s, it extremely is positive(b-m/n)?2) > a million. Equivalently 0 < a million/(ok?2) < b-m/n, subsequently a < m/n + a million/(ok?2) < b. The style m/n + a million/(ok?2) lies between a and b and is irrational through fact it extremely is the sum of the rational m/n and the irrational a million/(ok?2) .

Answer 3

Errr...

Yes ... or is it NO?