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a) prove that an integer n is a multiple of 9 if and only if the sum of its decimal digits is a multiple of 9. [Hint: Consider n − s, where s is the sum of the digits.]

b) Let n =
(765432101234567011223344556677007766554433221101234567)8.
(Thus n is expressed in octal notation.)
Is n a multiple of 7? [Hint: Use (i) generalized.]

please give some hint on how to do "both" of them.. thanks a lot for whoever that can help..

2007-09-26 01:25:28 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

for b, n=
(765432101234567011223344556677007766554433221101234567)8.

please follow this link and check the number. thanks so much~!!
http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/r/a2_2007.pdf

2007-09-26 01:26:55 · update #1

3 answers

a) for base b, prove lemma that a*b^j mod (b-1) = a
by a*b^j = a* (b^j - 1) + a and observe that b^j-1 is a string of digit (b-1)s and therefore divisible by b-1.
Now induce on the number of digits, m, of n. Write n as a*b^(m+1) + w. Use lemma and inducti0on hypothesis to calculate n mod (b-1) as a + sum of digits of w mod(b-1).

b)
(765432101234567011223344556677007766554433221101234567)
contains every octal digit the same number of times, so the sum of its digits is a multiple of 0+1+2+3+4+5+6+7 = ((0+7)+(1+6)+(2+5)+(3+4)) which is a multiple of 7.so the number itself is.

2007-09-26 09:50:10 · answer #1 · answered by holdm 7 · 0 0

a) When n is expressed in decimal notation, the representation means n = (a_k)*10^k + (a_(k-1))*10^(k-1) + ... + (a_1)*10 + (a_0). If we take congruences (mod 9), the powers of ten are all 1 (mod 9), so n = a_k + a_(k-1) + ... + a_1 + a_0 (mod 9). Now n is divisible by 9 iff n = 0 (mod 9), and the above shows that n in decimal notation and the sum of the digits are the same (mod 9).

b) Same idea, because the powers of 8 (in the octal notation) are all 1 (mod 7).

2007-09-26 01:40:31 · answer #2 · answered by Tony 7 · 0 0

Let be a number N=abc, what mean N= a*10^2 +b*10 +c

Let S be the sum of its decimal digits, i.e. S = a+b+c

Thus we consider the difference N -S = a* (10^2-1) +b*(10-1)+c-c = a*99+b*9 = 9*(11*a+b) that is multiple of 9, say
N-S = 9K

We have whenever N = (N-S) + S
N = (N-S) + S = 9K + S, that is multiple of 9 if and only if S is it.

2007-09-26 01:40:55 · answer #3 · answered by Filomates 4 · 0 1

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