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 Magic 8's (Posted on 2006-09-18)
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?

 No Solution Yet Submitted by joshua Rating: 4.0000 (2 votes)

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 solution | Comment 3 of 8 |

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+1100 wend `

 Posted by Charlie on 2006-09-18 15:51:46

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