Simple Divisor Problem - Math Puzzle
Source: http://www.math.utah.edu/~cherk/puzzles.html
Problem:
Prove that for any natural N, 1000^N - 1 cannot be a divisor of 1978^N - 1
Solution:
Highlight the part between the * symbols for the answer.
* Look at the power of 3 dividing each number. 1000^N - 1 always contains MORE factors of 3 than 1978^N - 1, so it can never divide it.
By the Lifting the Exponent lemma (LTE) for the odd prime 3: if 3 | a - 1 then v_3(a^N - 1) = v_3(a - 1) + v_3(N).
For 1000: 1000 - 1 = 999 = 27 x 37, so v_3(1000^N - 1) = 3 + v_3(N).
For 1978: 1978 - 1 = 1977 = 3 x 659 with 3 not dividing 659, so v_3(1978^N - 1) = 1 + v_3(N).
Hence v_3(1000^N - 1) = 3 + v_3(N) > 1 + v_3(N) = v_3(1978^N - 1) for every N. A divisor cannot contain a higher power of a prime than the number it divides, so 1000^N - 1 never divides 1978^N - 1. (Elementarily: 1000^N - 1 is a string of 3N nines, divisible by 27 as soon as... the digit-sum factorization 9 x 111 x 1001001 x ... pulls out one extra factor of 3 for each factor of 3 in N, always staying two ahead of 1978^N - 1.)
Solution by JDGM from the comments. *
Problem:
Prove that for any natural N, 1000^N - 1 cannot be a divisor of 1978^N - 1
Short and Sweet :)
Solution:
Highlight the part between the * symbols for the answer.
* Look at the power of 3 dividing each number. 1000^N - 1 always contains MORE factors of 3 than 1978^N - 1, so it can never divide it.
By the Lifting the Exponent lemma (LTE) for the odd prime 3: if 3 | a - 1 then v_3(a^N - 1) = v_3(a - 1) + v_3(N).
For 1000: 1000 - 1 = 999 = 27 x 37, so v_3(1000^N - 1) = 3 + v_3(N).
For 1978: 1978 - 1 = 1977 = 3 x 659 with 3 not dividing 659, so v_3(1978^N - 1) = 1 + v_3(N).
Hence v_3(1000^N - 1) = 3 + v_3(N) > 1 + v_3(N) = v_3(1978^N - 1) for every N. A divisor cannot contain a higher power of a prime than the number it divides, so 1000^N - 1 never divides 1978^N - 1. (Elementarily: 1000^N - 1 is a string of 3N nines, divisible by 27 as soon as... the digit-sum factorization 9 x 111 x 1001001 x ... pulls out one extra factor of 3 for each factor of 3 in N, always staying two ahead of 1978^N - 1.)
Solution by JDGM from the comments. *
The short and sweet answer is that 1000ᴺ-1 always has a higher power of three in its prime factorisation than does 1978ᴺ-1, so the former can never be a divisor of the latter.
ReplyDeleteA longer version goes something like this:
1000ᴺ-1 is a 3N-length string of 9s, so always has factors 9 and 111, and is thus always divisible by 27.
1978ᴺ-1 is only divisible by 27 when N = 9n.
However, when N = 9n, 1000ᴺ-1 is divisible by 243.
1978ᴺ-1 is only divisible by 243 when N = 81n.
But when N = 81n, 1000ᴺ-1 is divisible by 2187.
And so on, and in general:
3ʲ | 1978ᴺ-1 ⇒ N = 3ʲ⁻¹n ⇒ 3ʲ⁺² | 1000ᴺ-1
To understand why N = 3ʲn ⇒ 3ʲ⁺³ | 1000ᴺ-1, consider how factors with digit sum 3 are pulled out as N increases:
9 × 111
9 × 111 × 1001001
9 × 111 × 1001001 × 1000000001000000001
...
As for 3ʲ | 1978ᴺ-1 ⇒ N = 3ʲ⁻¹n, I do not actually have an explicit proof but believe it can be proved from elementary results and half a cup of coffee.
The straightforwardness seems plausible from the fact that Wolfram Alpha can calculate it so easily, even for biggish numbers and, at least at time of writing, has a bug for the case N = 1 that it gives no integer solutions to 1978ˣ (mod 3) = 1.
http://tech.groups.yahoo.com/group/mathforfun/message/10836
ReplyDeletecan we prove by using induction?
ReplyDeleteConsider 1978^N-1. For N=4i+1, one's digit is 7. For 4i+2, it is 3. 4i+3, it is 1. 4i, it is 5. And consider 1000^N-1 which has only either 8 or 0 in the one's place. They never match. Hence proved.
ReplyDelete???? 1000^N-1 is of the form 999 or 999999 or 999999999 .... and not as you have described. Actually I also worked on those lines but the solution would not lie there.
DeleteSolution is there in in the link http://tech.groups.yahoo.com/group/mathforfun/message/10836 using group theory .
ReplyDeleteClickable: http://tech.groups.yahoo.com/group/mathforfun/message/10836.
ReplyDeleteVivek: That link is spot on. Thanks. Everything is done "properly". My comment above has the right idea but the details are a bit dodgy, hehe.
Litu: My instinct is that this is one of those problems where the Math you would need for the inductive step is the same that could be used for the general case more easily.
Srikar: Firstly 1000ᴺ-1 always ends in 9, not "either 8 or 0". Secondly, if I understand you correctly then the premise of your proof is mistaken - a number can divide another number even if their final digits do not match.
Right idea, and the "Lifting the Exponent" lemma makes it airtight: v_3(1000^N - 1) = v_3(999) + v_3(N) = 3 + v_3(N), while v_3(197
ReplyDeleteThanks - (note for readers: Yahoo Groups shut down in 2020, so that link is now dead; the 3-adic / LTE argument in JDGM's comment ab
ReplyDeleteYou can, but the induction you would need is exactly the proof of the Lifting the Exponent lemma: v_3(x^(3N) - 1) = v_3(x^N - 1) +
ReplyDeleteTwo problems, as Vivek and JDGM noted: (1) 1000^N - 1 is 999...9, ending in 9 - not 8 or 0; (2) divisibility never required matchin
ReplyDeleteBoth corrections to Srikar stand: 1000^N - 1 is a string of 9s, and last-digit matching is not a divisibility criterion. And agreed
ReplyDeleteRight - 1000^N - 1 = 999...9 (3N nines), so any argument has to work with the 3-adic structure, which is what the group-theory
ReplyDelete