[1000 / 7 + 1000 / 7^2 + 1000 / 7^3] =142 + 20 + 2 = 164 [Take the integer part only].
So, the largest n = 164
+27 Here's (b)....it's pretty much an extension of (a)
1000/7 = 142 numbers that have at least 1 factor of 7
1000/49 = 20 numbers that have at an additional factor of 7
1000/343 = 2 numbers that have a third factor of 7
So......... 142 + 20 + 1 = 163
And 7^163 will divide 1000! ...so n = 163
+118608 1000/7 = 142.8571428571428571 floor = 142
flor(142/7) = 20
floor(20/7)=2
So I think 142+20+2=164
This is the same as the other two answers.
(afton just made a careles sum error at the end)
Factoring of 1000! gives:
1000!= 2^994 * 3^498 * 5^249 * 7^164 * 11^98 * 13^81 * 17^61 * 19^54 * 23^44 * 29^35 * 31^33 * 37^27 * 41^24 * 43^23 * 47^21 * 53^18 * 59^16 * 61^16 * 67^14 * 71^14 * 73^13 * 79^12 * 83^12 * 89^11 * 97^10 * 101^9 * 103^9 * 107^9 * 109^9 * 113^8 * 127^7 * 131^7 * 137^7 * 139^7 * 149^6 * 151^6 * 157^6 * 163^6 * 167^5 * 173^5 * 179^5 * 181^5 * 191^5 * 193^5 * 197^5 * 199^5 * 211^4 * 223^4 * 227^4 * 229^4 * 233^4 * 239^4 * 241^4 * 251^3 * 257^3 * 263^3 * 269^3 * 271^3 * 277^3 * 281^3 * 283^3 * 293^3 * 307^3 * 311^3 * 313^3 * 317^3 * 331^3 * 337^2 * 347^2 * 349^2 * 353^2 * 359^2 * 367^2 * 373^2 * 379^2 * 383^2 * 389^2 * 397^2 * 401^2 * 409^2 * 419^2 * 421^2 * 431^2 * 433^2 * 439^2 * 443^2 * 449^2 * 457^2 * 461^2 * 463^2 * 467^2 * 479^2 * 487^2 * 491^2 * 499^2 * 503 * 509 * 521 * 523 * 541 * 547 * 557 * 563 * 569 * 571 * 577 * 587 * 593 * 599 * 601 * 607 * 613 * 617 * 619 * 631 * 641 * 643 * 647 * 653 * 659 * 661 * 673 * 677 * 683 * 691 * 701 * 709 * 719 * 727 * 733 * 739 * 743 * 751 * 757 * 761 * 769 * 773 * 787 * 797 * 809 * 811 * 821 * 823 * 827 * 829 * 839 * 853 * 857 * 859 * 863 * 877 * 881 * 883 * 887 * 907 * 911 * 919 * 929 * 937 * 941 * 947 * 953 * 967 * 971 * 977 * 983 * 991 * 997
+26367 What is the largest integer n such that \(7^n\) divides 1000!
Alternate form
One may also reformulate Legendre's formula in terms of the base-p expansion of m.
Let s_{p}(m) denote the sum of the digits in the base-p expansion of m.
Source see: https://en.wikipedia.org/wiki/Legendre%27s_formula
Writing m = 1000 in binary as \(1000_{10} = 2626_{7}\),
we have that \(s_{7}(1000)=2+6+2+6=16\) and so
the largest integer n is \(\mathbf{\dfrac{m-s_{p}(m)} {p-1}}\)
\(\begin{array}{|rcll|} \hline n &=& \dfrac{m-s_{p}(m)} {p-1} \quad | \quad m=1000,\ p = 7,\ s_{7}(1000) = 16 \\\\ &=& \dfrac{1000-s_{7}(1000)} {7-1} \\\\ &=& \dfrac{1000-16} {7-1} \\\\ &=& \dfrac{984} {6} \\\\ \mathbf{n} &=& \mathbf{164} \\ \hline \end{array}\)
