Problems & Puzzles: Puzzles

Puzzle 568.- SOPF(N)=SOPF(RN)

JM Bergot sent the following puzzle:

One notices that 45 has a sum of factors 3+3+5=11; reverse the order of the digits to get 54, which also has a sum of factors= 3+3+3+2=11.

Q.  Send your largest example?

Contributions came from Carlos Rivera, Seiji Tomita, Torbjörn Alm, Giovanni Resta, J. K. Andersen, Ken Davis, Farideh Firoozbakht, Emmanuel Vantieghem, Jan van Delden.

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Carlos Rivera wrote:

I decided to search solutions where N -and consequently RN- is pandigital.

I found only one solution for pandigitals using numbers 1 to 9, and five solutions for pandigitals  using numbers 0 to 9.

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Seiji Tomita wrote:

Example of 100 digits.

N=1456666666666666666666666666666666666666666666666666666666666666666666666
  666666666666666666666666521

RN=1256666666666666666666666666666666666666666666666666666666666666666666666
  666666666666666666666666541

SOPF(N)=3+19+23+12004721+846035731396919233767211537899097169+10939984685537
       0537540339266842070119107662296580348039
      =109399846855370538386374998238989352874873834491449974
SOPF(RN)=3+29+13+12004721+846035731396919233767211537899097169+10939984685537
        0537540339266842070119107662296580348039
       =109399846855370538386374998238989352874873834491449974

SOPF(N)=SOPF(RN)

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Torbjorn Alm wrote:

My best solution

k= 9804997189 : 13+233+569+5689 = 6504
kr= 9817994089 : 19+139+653+5693 = 6504

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Giovanni Resta wrote:

A trick, which easily produce large examples,
is based on this observation:
if d = sopf(rev(n))-sopf(n) is divisible by 7 then
for m=n*10^(d/7) sopf(m)=sopf(n)+(d/7)*(2+5)=sopf(n)+d and
sopf(rev(m))=sopf(rev(n))=sopf(n)+d.
For example, sopf(25)=10 and sopf(52)=10+7,
hence sopf(250) = sopf(52).

A larger example, with 47,617,937 digits, is given
by m = 112089999*10^47617928, where
m =  2^47617928 x 3 x 5^47617928 x 7 x 97 x 1123  and
rev(m) = 3 x 333326727 and
2*47617928 + 3 + 5*47617928 + 7 + 97 + 1123 = 3 + 333326727.

I have played a little with a subproblem: to find n
such that sopf(n)=sopf(rev(n)) and the prime factors
of n and rev(n) have no common digits.
The smallest such number is
36679 = 43 x 853, where 97663 = 127 x 769,
and a larger such numbers are  33115520 and 97600000.

We can summarize this property with 3 observations:

1) sopf(x * y) = sopf(x) + sopf(y)
2) sopf(10^k) = 7*k
3) rev(x * 10^k) = rev(x) if x does not end in 0.

For another example:
Note that sopf(976) = sopf(2^4x61)=8+61=69 and sopf(679) = sopf(7x97)=7+97=104 and that (104-69) / 7 = 5.
Then we can add 5 zeros at 976 and
obtain a number, n=97600000, such that sopf(n)=sopf(rev(n)).
Since sopf(97600000)=sopf(976)+5*7 = 104 and
sopf(rev(97600000)) = sopf(rev(976)) = sopf(679)= 104.

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J. K. Andersen wrote:

Numbers with this property are in http://oeis.org/A085607: "Non-palindromic n
and its digit reversal have the same sum of prime factors (with repetition)"
45, 54, 250, 495, 594, 1131, 1311, 2262, 2550, 2622, 2750, 2926, 3393, ...

A085607 allows numbers ending in 0, for example 250 with reversal 052 = 52.
Assume an arbitrary number of ending and leading 0's are allowed in N and RN.
Let M be any number satisfying SOPF(M) = SOPF(RM)-7d for a positive integer d.
Let N = M*10^d. Then RN = RM, and N is a solution. Proof:
SOPF(N) = SOPF(M*2^d*5^d) = SOPF(M)+2d+5d = SOPF(M)+7d = SOPF(RM) = SOPF(RN)

The above can be used to generate enormous solutions when d is large.
In puzzle 20 I found the reversible prime
p = 10^2003-1+(689715601-999999999)*10^997.
Let M = reverse(p) = 10^2003-893482013*10^997-1.
M and RM=p are proven primes so SOPF(M) = M, and SOPF(RM) = RM.
SOPF(M) = SOPF(RM)-7d for d = (RM-M)/7 = 83313945*10^997.
d is a positive integer so this gives the solution
N = M*10^d = (10^2003-893482013*10^997-1)*10^(83313945*10^997).

N has 2003+83313945*10^997 digits including the ending 0's.
It would also be possible to find far larger solutions, especially if we
allow probable primes when SOPF(M) and SOPF(RM) are computed.


Now assume that ending and leading 0's are not allowed.
The palindromic prime p = 10^78942+10111100100111101*10^39463+1 was found by
Raffi Chaglassian in 2004:  http://primes.utm.edu/primes/page.php?id=72107
Let N = 1311*p. Then RN = 1131*p, and SOPF(N) = SOPF(RN) = p + 45,
since SOPF(1311) = SOPF(1131) = 45. N has 78946 digits.

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Ken Davis wrote:

It’s easy to show that any number of the form (45) repeated no matter what the length, is a solution.

Moreover:

it’s easy to show that any number consisting of a repeating pattern of an existing solution, no matter what the length, is a solution. eg 8749-8749-8749-....

Proof: I’ll use [] i.e. [N] = N repeating.

Let N be such that SOPF(N) = SOPF(RN)
Let X= [N]/N = [RN]/RN

Thus [N]= XN  and [RN] = XRN then

SOPF([N]) = SOPF(N) + SOPF(X)
SOPF([RN]) = SOPF(RN) + SOPF(X)

Then the fact that SOPF(N) = SOPF(RN) ==> SOPF([N]) = SOPF([RN])

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Farideh Firoozbakht wrote:

There exists infinitely many such numbers. In fact we have the following interesting

 theorem.

 

Theorem : If m is a number such that sop(m) = sop(reversal(m)) and 10 doesn't divide

m then for each positive integer n the number f(m,n) = m.m. ... .m (n times m)  has the

same property.

 

Proof : We must show that sop(f(m,n)) = sop(f(reversal(m).n)), for proving this We use of

these three facts.

 

1. If a & b be two positive integers then we have sop(a*b) = sop(a)+sop(b) (*) .

2. Since 10 doesn't divides m, number of digits of reversal(m) is equal to number of

digits of m.

3. For each number m we have f(m, n) = m*(10^(l*n) - 1)/(10^l - 1) where l is number of

digits of m.

 

 So,

  sop(f(m, n)) = sop(m*(10^(l*n) - 1)/(10^l - 1))

                   = sop(m) + sop((10^(l*n) - 1)/(10^l - 1))

                   = sop(reversal(m)) + sop((10^(l*n) - 1)/(10^l - 1))

                   = sop(reversal(m)*(10^(l*n) - 1)/(10^l - 1))

                   = sop(f(reversal(m), n)).

 

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Example : m = 45 & n = 5 so l = 2 and f(45, 5) = 4545454545, so

sop(4545454545) = sop(45*(101010101))

                          = sop(45) + sop(101010101)

                          = sop(54) + sop(101010101)

                          = sop(54*(101010101))

                          = sop(5454545454).

 

So by using the number 45, we can find arbitrary large numbers of even length which

have the same property. Namely all numbers of the form f(45,n) = 45.45. ... .45 (n times).

 

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Also for each integer k, 1<k<17  I found a(k), the smallest non-palindromic k-digit number m

with the mentioned property. a(2), a(3), ... & a(16) are respectively  :

 

45, 250, 1131, 12441, 109416, 1002921, 10009577, 100022593, 1000081008, 10000401424,

100000835544, 1000001449713, 10000013519782, 100000013605380 & 1000000081310530

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Emmanuel Vantieghem wrote:

Here are my "champions" for puzzle 568 :
10,000,046,130,534 = 2*3*3*43*1093*3361*3517  with SOP 8022
43,503,164,000,001 = 3*3*593*733*3083*3607  also with SOP 8022.

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Jan van Delden wrote:

1000001449713 & 3179441000001  with SOPF=3+7+17+1171+1483+1613=3+7+139+461+827+2857=429

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