Too many digits

Computer Science Level 2

Find the smallest positive integer $n$ for which the number of digits of $n!$ is not equal to the number of digits of $\displaystyle \sum_{k=1}^n k!$ .

Submit your answer as $0$ if you think that no such integer exists.

The answer is 96.

1 solution

Giorgos K.
Feb 20, 2018

You are wrong! Such an integer exists!
the smallest is 96

96!= 991677934870949689209571401541893801158183648651267795444376054838492222809091499987689476037000748982075094738965754305639874560000000000000000000000 has 150 digits

while Sum[n!,{n,96}]= 1002117819322478123365475165570275072629379236635019813428200288189501655552695033695642669679677899357177158363650532747296788236442336528920420940313 has 151 digits

Here is a Mathematica code that finds such integers
Select[Range@2000,IntegerLength[#!] != IntegerLength@Sum[k!, {k, #}]&]

the integers are 96, 261, 1556...

I found the same answer.

Ricardo Moritz Cavalcanti - 3 months, 2 weeks ago

Too many digits

The answer is 96.

1 solution

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