During Nov1-7, 2025, contriburions came from Paul Cleary

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Paul wrote:

Q2.  I couldn't find any bigger number than the 74 digits, Lexicographically using the 37 pairs it is the biggest number.

Q3.  I can confirm that it is the largest prime that can be built from those 37 pairs of numbers.


For completeness:-

this is the largest 70 digit number I could find

9999999997858585848484847673737373737373626262626262515151515140404040
and the 70 digit prime is
9999999997858585848484847673737373737373626262626262404051515151404051

this is the largest 72 digit number I could find

999999989898868684848484757573737373737373626262626262515151515140404040
and there is no prime with this set of digit pairs, the sum of digits is 369

Why a 76 digit number doesn't exist.

What these numbers are,
The number is built from 2-digit blocks like “73”, “95”, “40”, and so on.
Each block “ab” means: digit b appears a times in the whole big number.
Since the overall length is even, a 76-digit example would have 38 such blocks.

Two basic facts

A block can’t start with 0. If you wrote “0b” you’d be claiming “digit b appears 0 times” while you are literally using b in that block. So every zero in the whole number must be the last digit of some block like “…0”.

The count for each digit must be consistent. If the number says “40” anywhere, that means “there are 4 zeros”. You can’t also say “30” or “50”. Every block that ends with 0 must be “40”, and the total number of such “40” blocks must equal the number of zeros.

This forces: 0→4, 1→5, 2→6, 3→7
In the large valid examples, the four smallest digit counts settle at

0 appears 4 times,
1 appears 5 times,
2 appears 6 times,
3 appears 7 times.

Why this is natural: the only way to supply all zeros is with copies of “40”, and because there are exactly 4 zeros you must have exactly four “40” blocks. The same consistency then cascades upward and fixes “51” five times, “62” six times, and “73” seven times.  These four blocks together account for 4+5+6+7=22 digits of the 76.

What 76 digits would force on the big digits

The ten digit counts must add up to 76. After the 22, there are 54 appearances left to split among the six larger digits 4, 5, 6, 7, 8, 9. Each count is a single digit (0 to 9). The only way to reach a total of 54 with six one-digit numbers is that each of 4, 5, 6, 7, 8, 9 appears exactly 9 times.
So at 76 digits, the target counts would be:


0→4, 1→5, 2→6, 3→7, and 4→9, 5→9, 6→9, 7→9, 8→9, 9→9.

Lets just focus on just one digit: the 4s.
We need nine 4s altogether. A 4 can appear as:

the tens digit of “4b”, which states “digit b appears 4 times”, or
the units digit of “a4”, which states “digit 4 appears a times”.

But which “4b” blocks are allowed? Only the digit 0 appears 4 times, so the only “4b” we can use is “40”. We already know there are exactly four “40” blocks, one for each zero. That supplies four of the needed 4s (they sit at the tens place in those four “40” blocks). We still need five more 4s. The only way to add 4s now is with blocks that end in 4, and since digit 4 must appear 9 times, those must be “94” blocks. So we are forced to include five “94” blocks.  Now count how many 9s you must spend,
every block that starts with 9 contributes a 9 in the tens place. We just saw we already need five “94” blocks. Do the same minimal bookkeeping for the other digits:

To make five 1s you need five “51” blocks, which still leaves four more 5s to reach nine, so you need four “95” blocks.  To make six 2s you need six “62” blocks, which still leaves three more 2s to reach nine 2s at tens, so you need three “96” blocks.  To make seven 3s you need seven “73” blocks, which still leaves two more 3s, so you need two “97” blocks.  No digit occurs exactly 8 times, so the only way to make nine 8s is nine “98” blocks.

Add up the blocks that start with 9 so far:
“94” five times,
“95” four times,
“96” three times,
“97” two times,
“98” nine times.

That totals 5 + 4 + 3 + 2 + 9 = 23 blocks starting with 9.
The whole 76-digit number is only allowed nine 9s in total, because the count for digit 9 must be 9. Yet we have already shown we would need at least 23 blocks that start with 9, which would place at least 23 9s at tens positions alone. That blows the “nine 9s” budget long before we even think about any “99” blocks.

Conclusion

The 76-digit case forces all of 4, 5, 6, 7, 8, 9 to appear nine times. That in turn obliges you to use far more blocks that start with 9 than the nine available 9s can support. You run out of 9s. Since the counts must be single digits and globally consistent, there is no way around this. Therefore a 76-digit number cannot exist.

Here are the digit pairs for length 70, 72, and 74

L = 70
Pairs:

99 × 4
97 × 1
85 × 3
84 × 4
76 × 1
73 × 7
62 × 6
51 × 5
40 × 4

L = 72
Pairs:

99 × 3
98 × 3
86 × 2
84 × 4
75 × 2
73 × 7
62 × 6
51 × 5
40 × 4

L = 74
Pairs:

99 × 1
98 × 2
97 × 2
96 × 3
85 × 3
84 × 4
73 × 7
62 × 6
51 × 5
40 × 4

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