Finding a number that has high multiplicative persistence is a prime example of "hard to find, easy to verify". Given a (say) 500-digit number, it only takes a few lines of code, and an eyeblink of computer time, to compute its persistence.

My general advice in such situations is:

If you think you have succeeded in the difficult "find" task, surely you should spend a reasonable effort in the relatively easy "verify" task before announcing your result.

So, if you think you have found such a small number with persistence 12, surely you should first verify your claim with some simple computer code. If that seems to succeed, perhaps double-check using another implementation, perhaps in a different language, perhaps on a different computer. After a reasonable amount of verification succeeds -- OK, perhaps then it is time to think about telling others.

Some basic references on multiplicative persistence may be helpful: MathWorld, Wikipedia, OEIS A003001. The comments in OEIS claim various lower bounds on the size of a number of persistence 12: a comment in 2019 claims a lower bound of $10^{20585}$, and a comment in 2023 claims $2.67 \times 10^{30000}$. A 500-digit number would contradict both, so all the more reason to double-check.

For completeness, here is straightforward SageMath code to verify such claims. It can be run online in this SageMathCell.

def digits(n):
    d = []
    while n:
        n, remainder = divmod(n, 10)
        d.append(remainder)
    return d
        
def multiplicative_persistence(n):
    i = 0
    print("%d: %d" % (i,n))
    while n>=10:
        n = prod(digits(n))
        i += 1
        print("%d: %d" % (i,n))
    print("It has multiplicative persistence %d\n" % (i))

multiplicative_persistence(277777788888899)
multiplicative_persistence(10^500-1)

The output from these numbers is:

0: 277777788888899
1: 4996238671872
2: 438939648
3: 4478976
4: 338688
5: 27648
6: 2688
7: 768
8: 336
9: 54
10: 20
11: 0
It has multiplicative persistence 11

0: 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
1: 1322070819480806636890455259752144365965422032752148167664920368226828597346704899540778313850608061963909777696872582355950954582100618911865342725257953674027620225198320803878014774228964841274390400117588618041128947815623094438061566173054086674490506178125480344405547054397038895817465368254916136220830268563778582290228416398307887896918556404084898937609373242171846359938695516765018940588109060426089671438864102814350385648747165832010614366132173102768902855220001
2: 0
It has multiplicative persistence 2

First, to your direct question: "Is the record still 11", the answer is "yes", according to a variety of sources (MathWorld, Wikipedia, OEIS A003001). If anyone has found a new record, at least it is not in these sources yet.

Now for some general advice.

Finding a number that has high multiplicative persistence is a prime example of "hard to find, easy to verify". Given a (say) 500-digit number, it only takes a few lines of code, and an eyeblink of computer time, to compute its persistence.

My advice in such situations is:

If you think you have succeeded in the difficult "find" task, surely you should spend a reasonable effort in the relatively easy "verify" task before announcing your result.

So, if you think you have found such a small number with persistence 12, surely you should first verify your claim with some simple computer code. If that seems to succeed, perhaps double-check using another implementation, perhaps in a different language, perhaps on a different computer. After a reasonable amount of verification succeeds -- OK, perhaps then it is time to think about telling others.

The comments in OEIS claim various lower bounds on the size of a number of persistence 12: a comment in 2019 claims $10^{20585}$, and a comment in 2023 claims $2.67 \times 10^{30000}$. A 500-digit number would contradict both, so all the more reason to double-check.

For completeness, here is straightforward SageMath code to verify such claims. It can be run online in this SageMathCell.

def digits(n):
    d = []
    while n:
        n, remainder = divmod(n, 10)
        d.append(remainder)
    return d
        
def multiplicative_persistence(n):
    i = 0
    print("%d: %d" % (i,n))
    while n>=10:
        n = prod(digits(n))
        i += 1
        print("%d: %d" % (i,n))
    print("It has multiplicative persistence %d\n" % (i))

multiplicative_persistence(277777788888899)
multiplicative_persistence(10^500-1)

The output from these numbers is:

0: 277777788888899
1: 4996238671872
2: 438939648
3: 4478976
4: 338688
5: 27648
6: 2688
7: 768
8: 336
9: 54
10: 20
11: 0
It has multiplicative persistence 11

0: 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
1: 1322070819480806636890455259752144365965422032752148167664920368226828597346704899540778313850608061963909777696872582355950954582100618911865342725257953674027620225198320803878014774228964841274390400117588618041128947815623094438061566173054086674490506178125480344405547054397038895817465368254916136220830268563778582290228416398307887896918556404084898937609373242171846359938695516765018940588109060426089671438864102814350385648747165832010614366132173102768902855220001
2: 0
It has multiplicative persistence 2

Finding a number that has high multiplicative persistence is a prime example of "hard to find, easy to verify". Given a (say) 500-digit number, it only takes a few lines of code, and an eyeblink of computer time, to compute its persistence.

My general advice in such situations is:

If you think you have succeeded in the difficult "find" task, surely you should spend a reasonable effort in the relatively easy "verify" task before announcing your result.

So, if you think you have found such a small number with persistence 12, surely you should first verify your claim with some simple computer code. If that seems to succeed, perhaps double-check using another implementation, perhaps in a different language, perhaps on a different computer. After a reasonable amount of verification succeeds -- OK, perhaps then it is time to think about telling others.

Some basic references on multiplicative persistence may be helpful: MathWorld, Wikipedia, OEIS A003001. The comments in OEIS claim various lower bounds on the size of a number of persistence 12. It may be helpful to consider those comments.

For completeness, here is straightforward SageMath code to verify such claims. It can be run online in this SageMathCell.

def digits(n):
    d = []
    while n:
        n, remainder = divmod(n, 10)
        d.append(remainder)
    return d
        
def multiplicative_persistence(n):
    i = 0
    print("%d: %d" % (i,n))
    while n>=10:
        n = prod(digits(n))
        i += 1
        print("%d: %d" % (i,n))
    print("It has multiplicative persistence %d\n" % (i))

multiplicative_persistence(277777788888899)
multiplicative_persistence(10^500-1)

The output from these numbers is:

0: 277777788888899
1: 4996238671872
2: 438939648
3: 4478976
4: 338688
5: 27648
6: 2688
7: 768
8: 336
9: 54
10: 20
11: 0
It has multiplicative persistence 11

0: 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
1: 1322070819480806636890455259752144365965422032752148167664920368226828597346704899540778313850608061963909777696872582355950954582100618911865342725257953674027620225198320803878014774228964841274390400117588618041128947815623094438061566173054086674490506178125480344405547054397038895817465368254916136220830268563778582290228416398307887896918556404084898937609373242171846359938695516765018940588109060426089671438864102814350385648747165832010614366132173102768902855220001
2: 0
It has multiplicative persistence 2

Finding a number that has high multiplicative persistence is a prime example of "hard to find, easy to verify". Given a (say) 500-digit number, it only takes a few lines of code, and an eyeblink of computer time, to compute its persistence.

My general advice in such situations is:

If you think you have succeeded in the difficult "find" task, surely you should spend a reasonable effort in the relatively easy "verify" task before announcing your result.

So, if you think you have found such a small number with persistence 12, surely you should first verify your claim with some simple computer code. If that seems to succeed, perhaps double-check using another implementation, perhaps in a different language, perhaps on a different computer. After a reasonable amount of verification succeeds -- OK, perhaps then it is time to think about telling others.

Some basic references on multiplicative persistence may be helpful: MathWorld, Wikipedia, OEIS A003001. The comments in OEIS claim various lower bounds on the size of a number of persistence 12: a comment in 2019 claims a lower bound of $10^{20585}$, and a comment in 2023 claims $2.67 \times 10^{30000}$. A 500-digit number would contradict both, so all the more reason to double-check.

For completeness, here is straightforward SageMath code to verify such claims. It can be run online in this SageMathCell.

def digits(n):
    d = []
    while n:
        n, remainder = divmod(n, 10)
        d.append(remainder)
    return d
        
def multiplicative_persistence(n):
    i = 0
    print("%d: %d" % (i,n))
    while n>=10:
        n = prod(digits(n))
        i += 1
        print("%d: %d" % (i,n))
    print("It has multiplicative persistence %d\n" % (i))

multiplicative_persistence(277777788888899)
multiplicative_persistence(10^500-1)

The output from these numbers is:

0: 277777788888899
1: 4996238671872
2: 438939648
3: 4478976
4: 338688
5: 27648
6: 2688
7: 768
8: 336
9: 54
10: 20
11: 0
It has multiplicative persistence 11

0: 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
1: 1322070819480806636890455259752144365965422032752148167664920368226828597346704899540778313850608061963909777696872582355950954582100618911865342725257953674027620225198320803878014774228964841274390400117588618041128947815623094438061566173054086674490506178125480344405547054397038895817465368254916136220830268563778582290228416398307887896918556404084898937609373242171846359938695516765018940588109060426089671438864102814350385648747165832010614366132173102768902855220001
2: 0
It has multiplicative persistence 2