Question

For which $n$ is the sum of digits of $n!$ equal to 9? Recall. $n! = n(n-1)...3\cdot2\cdot1$, and as an example: $10! = 3628800$ and its sum of digits is $3 + 6 + 2 + 8 + 8 + 0 + 0 = 27 \neq 9$, so 10 is not one of the numbers we are looking for.

          For which $n$ is the sum of digits of $n!$ equal to 9?
Recall. $n! = n(n-1)...3\cdot2\cdot1$, and as an example: $10! = 3628800$ and
its sum of digits is $3 + 6 + 2 + 8 + 8 + 0 + 0 = 27 \neq 9$, so 10 is not one of
the numbers we are looking for.

For which n is the sum of digits of n! equal to 9?
Recall. n! = n(n-1)...3·2·1, and as an example: 10! = 3628800 and
its sum of digits is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27 ≠ 9, so 10 is not one of
the numbers we are looking for.

Added by Mark M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Hunza Gilgit

01/09/2022

Step 1

Step 1: Start with n=1 and calculate the factorial and the sum of its digits. Show more…

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For which n is the sum of digits of n ! equal to 9 ? Recall. n!=n*(n-1)*dots*3*2*1, and as an example: 10!=3628800 and its sum of digits is 3+6+2+8+8+0+0=27!=9, so 10 is not one of the numbers we are looking for. For which n is the sum of digits of n! equal to 9? Recall. n!= n.(n -1).....3.2.1, and as an example: 10!=3628800 and its sum of digits is 3+6+2+8+8+0+0=279so 10 is not one of the numbers we are looking for.