How do I solve the following problem step by step? and what are the solution? I

Question

How do I solve the following problem step by step? and what are the solution? I am really lost, and do not understand the whole exercise..

Let a be a number written (in base 10) as

(i) Prove that 2 divides a if and only if 2 divides a0

(ii) Prove that 4 divides a if and only if 4 divides a0+2a1

(iii) Prove that 4 divides a if and only if 4 divides a0+2a1+4a2

(iv) Prove that 5 divides a if and only if 5 divides a0

(v) Prove that 9 divides a if and only if 9 divides the sum a0+a1+...+an of its digits.

(vi) Prove that 3 divides a if and only if 3 divides the sum of its digits

(viii) What is the rule for divisibility by 7?

Explanation / Answer

Don't really count this as an answer, because this is a really long, tedious question to fully answer. The basic idea is going back to the divisibility tricks you probably learned in 5th grade. If theres a number like 123023012392, you know its divisible by 2 because the last digit, the 2, is divisible by 2. Why? Because you know that 2*5 is 10, and for that matter, 2*50 is 100, and so on and so forth, so we know that 2 can divide any of the other digits, and we only haveto look at the last one. This is essentially what your problem is saying, because a0 * 10^0 is just a0 * 1 = a0. So this is the 1's digit. The same is true for a1 * 10, this makes a1 the tens digit, and so forth. Where it gets hard is when you have the harder divisibility rules to follow. 4 is still okay, look at what the multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40... The pattern repeats, and notice that the only time 4 does not divide just a0 is when the a1 digit, the tens digit, has an odd number. At these places, the a0 digit is shifted 2 lower than it normally is, and you would have to proceed for a rigorous proof with talking about how any odd number can be represented as 2*n+1 where n is an integer. This means that when you take 2 times that odd number, you get 4*n, which is divisible by 4 clearly, and then +2, which accounts for the offset of the a0 digit for these odd a1 digits. When a1 is even then, its easy enough as any even is 2*n, and 2 times that is just 4*n, so you're golden. The part 3 i can only assume is meant as a trick, since you already know a0 + 2a1 is divisible by 4, adding on 4*a2, which is also divisible by 4, must also make a number divisible by 4. You could show that a bit more rigorously by saying from part 2 a0+2a1 = 4*x, for some x in Z, so clearly a0+2a1+4a2 = 4*x + 4*a2 = 4*(x+a2) is also divisible by 4. part 4 is almost identically part 1, you know that 5*2 = 10, and 5*20 = 100, etc... so only the a0 digit matters. part 5 I don't think I can come up with a clever way of doing... I would probably throw together some kind of quasi-inductive proof about how it is true for all the multiples 9,18,27,36,....,81,90,99. Then show that past 99, you essentially shift the a0 and the a1 digits sum down by 1, looking at 108, 117, etc... It is basically 1 less than the 9, 18, ... we had before, because the 1 is in the 100's place. This continues on for the a2, hundreds place, until 999, when you do the exact same kind of shift, and so on and so forth. Theres probably a more clever & thorough way of proving it, though... part 6 I would defer to part 5, and basically prove that for 3,6,9,12,...30, then again for 33,36,39,42...,60 and lastly for 63,66,69,72,...90 you have this rule hold true. After that it is back to part 5 proving that the 99 pushing a digit over to 108 argument still works. part 7 - I frankly don't think this is true? Am I reading it wrong? I think this would say that 11 doesnt divide 66 because 6-6 = 0, and 11 doesn't divide 0... Same is true for say 11*22 = 242, we have 2 - 4 + 2 = -2... And 11 doesn't divide -2 either...? I think theres something wonky here. part 8 - this is also something I can't give you a straight answer on. You don't learn a divisibility rule for 7 normally, because its a very wonky number, there is no nice way to form 10, 100, 1000, 10000, ... from the number because at every step, the closest you get nicely is 7, 70+28, 700+280+14,... The closest thing to an answer I can find is http://en.wikipedia.org/wiki/Divisibility_rule I think that looking closely at the pattern of how the change in what you need to add to get to those 10,100,1000,10000,... numbers will give you the best chance, and it looks like the pattern you will get will be multiplying the digits by 1,3,2,-1,-3,-2,1,3,2,-1,-3,-2,... and so on. There might also be some promise in the remove the last digit, and subtract 2 * that last digit from what remains. But thats going to be about as much of a mess to prove. Hope that helps at least a bit, its a tough exercise, goodluck.

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