ccheng26 wrote:
Can someone explain this further? I'm not following the solutions here. Thanks!
OK so we are looking for maximum value of n when \(7^n<5000!\).
This basically means the number of times 7 comes in the product 1*2*3*4*...*5000.
Now when we divide by 7 what happens..
\(\frac{5000}{7}=714.xy\), so 714 gives us the number which have 7 in it.. 7, 14, 21, ......4998.
\(\frac{5000}{49}=102.04\)~120 gives us the number which are div by 49 as there is an extra 7 in these numbers... 49, 343....
\(\frac{5000}{49*7}=14.57\)~14 gives us the number which are div by 49*7 as there is an extra 7 in these numbers over and above what we have counted till now... 343....
similarly \(\frac{5000}{49*7*7}=2.08\), so 2 numbers more 2401 and 4802
Next 49*7*7*7 goes beyond 5000 and will give us an answer less than 1..
Now we have calculated all 7s whether as 49, 343 etc = 714+102+14+2=832
So our Ans 10+3+1=14
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