There is a number which has 8 divisors including, 8, itself, and 1.
There is also a number which has 18 divisors, including 18, itself and 1.
The difference between these numbers is 28.
What are the two numbers?
In order for a number to have n factors, its prime factorization, p1^k1 * p2^k2 * ..., must be such that (k1+1) * (k2+1) * ... = n, as, in forming a factor, there is a choice of zero through k1 occurrences of the first prime, zero through k2 occurrences of the second, etc.
In the case of needing 8 factors, including 2^3=8, the possibilities are 2^7 or 2^3 * p1. The first, 2^7=128, is eliminated by the fact that neither 128+28=156 nor 128-28=100 meets the criteria for the second number: The former is not divisible by 18, and has only 12 factors; the latter is also not divisible by 18 and has only 9 factors. So the first number is 8 * p1.
In order for the second number to have 18 factors, its prime factorization could be of the form p2^17, or p2^2*p3^5 or p2*p3^8 or p2*p3^2*p4^2. But it must also be divisible by 18=2*3^2, so at least the two primes, 2 and 3, are involved.
2^2*3^5 = 972; 28 less is 944, which has 10 factors rather than 8
28 more is 1000, which has 16 rather than 8.
3^2*2^5 = 288; 28 less is 260, which is not divisible by 8 and has 12 factors
; 28 more is 316, which is not divisible by 8 and has 6 factors.
2*3^8 = 13122; 28 less is 13,094, which is not divisible by 8 and has 4 factors
; 28 more is 13,150, which is not divisible by 8 and has 12 factors.
We can't reverse the positions of the 2 and the 3 in this form (i.e., have 3*2^8) as the power of 3 must be at least 2.
That leaves 2*3^2*p4^2 or p2*2^2*3^2, as the second number. However, since the first number is a multiple of 8, when we add or subtract 28 to/from it, that second number must be a multiple of 4, so the 2 must be squared, and the form is p2*2^2*3^2 = 36*p2.
Combining the above algebraically, we determine that 2*p1 differs from 9*p2 by 7. Also, p2 can't be 2 or 3 and p1 can't be 2, in order to meet the prime-factorization forms of the numbers. Trying the simplest p2 = 5, results in p1 being either (45+7)/2=26 or (45-7)/2=19. The former is not prime, but 19 is.
Remember the first number is 8*p1 and the second is 36*p2, so p2=5 means the second number is 5*36=180, and p1=19 gives the first number as 152. So this is a solution.
But we can try other numbers besides 5 as p2.
Here's a list of the first 35 successful results:
p2 first num second number
5 152 180
11 424 396
17 584 612
19 712 684
23 856 828
29 1016 1044
37 1304 1332
41 1448 1476
43 1576 1548
59 2152 2124
61 2168 2196
79 2872 2844
89 3176 3204
97 3464 3492
103 3736 3708
109 3896 3924
131 4744 4716
137 4904 4932
151 5464 5436
179 6472 6444
181 6488 6516
191 6904 6876
197 7064 7092
211 7624 7596
257 9224 9252
263 9496 9468
271 9784 9756
283 10216 10188
317 11384 11412
331 11944 11916
349 12536 12564
359 12952 12924
379 13672 13644
389 13976 14004
397 14264 14292
10 P2=3
20 while Ct<35
30 P2=nxtprm(P2)
40 P1=(9*P2-7)//2
50 if P1=int(P1) then if nxtprm(P1-1)=P1 then if P1<>2 then print using(6,0),P2,8*P1,8*P1+28:Ct=Ct+1
60 P1=(9*P2+7)//2
70 if P1=int(P1) then if nxtprm(P1-1)=P1 then if P1<>2 then print using(6,0),P2,8*P1,8*P1-28:Ct=Ct+1
100 wend
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Posted by Charlie
on 2006-09-18 15:51:46 |
solution
