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What I am trying to achieve, is related to cryptography/blockchain/bitcoin . So, the largest number here is huge, in other words: I want to find the largest multiple of 7, which is lower than this number:

$115792089237316195423570985008687907852837564279074904382605163141518161494336 $

I can just go to Wolfram Alpha, and type "multiples of 7", and I get a list of the multiples relatively fast. But, it will take some time until I keep hitting "more", to get to a number lower than this above.

One can compute this number $a$ modulo $7$. The result is $2bmod 7$. So take $a-2$. It is the largest multiple of $7$ less than $a$.

$$begin{array}{cccccc}115792&089237&316195&423570&985008&687907\852837&564279&074904&382605&163141&518161\494336end{array}$$
Sum up the places of these numbers, by place value carrying when needed, then apply $10^kequiv 3^k bmod 7$ you'll then have a much smaller number to find the remainder of that's equivalent.
5667972, which goes to :$$6(3^5)+6(3^4)+2(3^2)equiv 1458+486+18equiv 2+3+4equiv 2 bmod 7$$ so the largest multiple of 7, is 2 less than the number. Yes, this is a slightly tedious way to go, but it's inspired via extension of Fermat's little theorem, and polynomial remainder theorem.

The reason I broke it into 6 digits at a time, is because Fermat's extension, is that exponents that have the same remainder mod p-1, will give back the same remainder with the same base. That means you can simply turn one into the other, adding like terms. you then go and do the addition the first column on the right sums to 62, carry the 6, that means you sum the next column plus 6, giving 57 carry the 5, next column is then 59, carry the 5, next column 67, carry the 6, next column, 76 carry the 7, next column, 56 there's no column to carry the 5 onto, and in the next step, it will be merged with the 2 (6 digits before), and then tossed because 7 creates a term that is 0 mod 7. Doing the same to other 7's and the nine gives 660200 we then replace x=10 with 3, via polynomial remainder theorem, and evaluate the sum shown above.Formula used $$sum_{n=0}^Ld_na^nequivsum_{n=0}^L(d_nbmod p)(a_nbmod p)^{(n bmod (p-1))} pmod p$$ we did the exponent part first, the base part second, and the coefficient (digit) part third, we then used the simple reduction mod p last.

Just divide the number by 7, if the mod is 0, you subtract 1 from the quotient and multiply it by 7, else, the quotient times 7 is your desired number.

Ex:
70 / 7 = 10, with mod 0. 10-1 = 9 => 9 * 7 = 63 >Biggest multiple under 70.

71 / 7 = 10, with mod 1. 10 * 7 = 70 => Biggest multiple under 71