1×2×3×4×5×6×7×8×9×…… How about 99 × 100? Let's show what the unknowns mean. It's better to have a formula, otherwise we won't be able to solve other problems

1×2×3×4×5×6×7×8×9×…… How about 99 × 100? Let's show what the unknowns mean. It's better to have a formula, otherwise we won't be able to solve other problems

N * (n + 1) = n ^ 2 + N, so
Original formula = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 +. + n ^ 2 + 1 + 2 + 3 +... + n
＝n*(n+1)（2n＋1）/6+n*(1+n)/2
n=99
=99*100*199/6+99*100/2=328350+4950=333300
1*2*3*4*… *99*100
=100!
=9.3326215443944152681699238856267e+157
Where * is a multiplication sign and E + 157 is the 157 times of 10
This kind of calculation can't be worked out with a calculator,
You have to use the calculator in the computer,
The number is 9.332621543944152681699238856267e + 157
In addition, n! Denotes 1 * 2 * 3. * (n-1) * n,
If you want to find out how many zeros there are at the end of the result, the method is as follows:
Let's push backward
To know how many zeros there are, we actually need to know how many powers of 10 there are in the result
10 = 2 * 5. Obviously, from 1 to N, the number of 2 is much larger than that of 5
Actually, we need to find out how many powers there are 5 in the result, which is actually how many 5's there are
1) Suppose that the idempotent of 5 closest to n is the natural number to the power a of 5 (then the idempotent close to 100 is the power 2 of 5 = 25, a = 2)
2) The answer is a * (power a of N / 5) + (A-1) * (power a of N / 5-power A-1 of N / 5) + (power A-2 of N / 5-power A-1 of N / 5-power a of N / 5) +
So the answer to the question is:
2 * (100 / 25) + 1 * (100 / 5-100 / 25) = 8 + 16 = 24 zeros
How many zeros are there at the end of 1 * 2 * 3 *. * 688?
5 * 5 * 5 * 5 = 625, but the 5th power of 5 is more than 688
So, a = 4
The answer is:
4 * ([688 / 625]) + 3 * ([688 / 125] - [688 / 625]) + 2 * ([688 / 25] - [688 / 125] - [688 / 625]) + 1 * ([688 / 5] - [688 / 25] - [688 / 125] - [688 / 625]) = 4 * 1 + 3 * 3 + 2 * 23 + 110 = 169 zeros