| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
It has been observed that: 2^5 + 3^1 = 35 (a multiple of 5) 2^4 + 3^2 = 25 ( " ) 2^3 + 3^3 = 35 ( " ) 2^2 + 3^4 = 85 ( " ) Prove of disprove the following statement: If 2^a + 3^b (where a & b >0) is a multiple of 5, then 2^(a+1) + 3^(b-1) is also a multiple of 5. Further prove or disprove, if 2^a + 3^b (where a & b >0) is a multiple of 5, then (i) 2^(a+k) + 3^(b-k) is also a multiple of 5 (as long as (a+k) and (b-k) >0); (ii) 2^(a-k) + 3^(b+k) is also a multiple of 5 (as long as (a-k) and (b+k) >0). Regards, Simon
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 549.1 | CLT::GILBERT | eager like a child | Thu Jul 31 1986 08:05 | 6 | |
and (iii) Find corresponding numbers that produce multiples of p, for any prime p. (iv) Find corresponding numbers that produce multiples of m, for any m, when such (distinct) numbers exist. | |||||
| 549.2 | Divisibility by 5 - A Generalisation | PRCSWS::SIMONLI | Fri Aug 01 1986 06:57 | 31 | |
Further, prove (or disprove), in general, that: (v) If (10m+2)^a + (10n+3)^b is a multiple of 5 (where a & b >0, and M & n >=0), then: the sum of two terms (10m+2)^(a+k) and (10n+3)^(b-k) is also a multiple of 5 (as long as this sum >0). That is, (10m+2)^(a+k) + (10n+3)^(b-k) is also a mutiple of 5. NOTE: (1) (ii) above is a case of (i) by substituting k=-k. (2) (i) is a general case of (v) (when m & n are both =0). (vi) The same holds if the sum in (v) above is replaced by the difference (of course, the difference has to be >0). That is, if (10m+2)^a - (10n+3)^b is a multiple of 5, then (10m+2)^(a+k) - (10n+3)^(b-k) is also a multiple of 5. For example, using 2 and 3 again (i.e., m=n=0): 2^8 - 3^4 = 175 2^9 - 3^3 = 485 2^10- 3^2 = 1015 (NOTE: even the difference <0, it is still a multiple of 5: 2^7 - 3^5 = -115 2^6 - 3^6 = -665 So, we may even further formulate this to be just the DIFFERENCE of the two terms, i.e., the larger term - the smaller term.) Regards, Simon Li | |||||
| 549.3 | CLT::GILBERT | eager like a child | Fri Aug 01 1986 12:27 | 5 | |
(v) (10m+2)^a + (10n+3)^b = 0 modulo 5 iff 2^a + 3^b = 0 modulo 5. Thus, a proof of (ii) gives a proof of (v). ( are these problems being ignored because they're too simple, or is everyone trying to think of ways to make the problems harder? ) | |||||
| 549.4 | CLT::GILBERT | eager like a child | Fri Aug 01 1986 18:27 | 1 | |
Hint: 2*3 = 1 modulo 5, so 2 and 3 are multiplicative inverses modulo 5. | |||||
| 549.5 | AURORA::HALLYB | God throws loaded dice | Sun Aug 03 1986 13:47 | 5 | |
Sometimes questions are phrased in such a manner that the author(s)
do not make it clear if the problem is (a) unsolved, or (b) simply
a fun exercise the reader might enjoy.
John
| |||||
| 549.6 | Does that matter? | 22537::SIMONLI | Mon Aug 18 1986 03:06 | 5 | |
Does that matter?
For this one, at first I did not have the solution when I posted
up this note, but then I found a way of solving it.
| |||||
| 549.7 | Outline of proof | ZFC::DERAMO | My very own personal name | Thu Dec 10 1987 19:18 | 27 |
Spoiler.
Assume a > 0, b > 1
Start with this identity: 3^b = 3 * 3^(b-1)
Multiply both sides by 2: 2 * 3^b = 6 * 3^(b-1)
>> .4
>> Hint: 2*3 = 1 modulo 5, so 2 and 3 are multiplicative inverses modulo 5.
Reduce modulo 5: 2 * 3^b == 1 * 3^(b-1) mod 5
2 * 3^b == 3^(b-1) mod 5
Add in the following identity: 2 * 2^a = 2^(a+1)
The sum of the last two: 2 * (2^a + 3^b) == 2^(a+1) + 3^(b-1) mod 5
Assume 2^a + 3^b == 0 mod 5: 0 == 2^(a+1) + 3^(b-1) mod 5
So assuming 2^a + 3^b is divisible by 5, where a > 0 and b > 1,
implies 2^(a+1) + 3^(b-1) is also divisible by 5.
Reverse the roles of 2 and 3 for 2^(a-1) + 3^(b+1).
Use induction to extend these from +/- 1 to +/- k.
Dan
| |||||