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Number Theory Level 3

Find the last digit of $(1!+2!+\cdots+1729!)^{1729}$ .

This problem is part of Math's Planet

2 solutions

Yash Dev Lamba
Mar 3, 2016

$N = 1!+2!+3!+4!+5!+.....+1729!$

$=(1!+2!+3!+4!)+(5!+.....+1729!)$

$=33+$ an integer having $0$ in its unit's place

= an integer having $3$ in its unit's place.

Hence, $N^{1729}$ is an integer having $3$ in its unit's place.

Aww I didn't even think about 5!... so simple

Johnny Jillky - 5 years, 2 months ago

Ashwin K
Mar 6, 2016

Looks difficult but pretty easy problem :

We need last digit, for n>4, n! will end with 0 and so we need to consider only first 4 numbers for factorial.

1! + 2! + 3! + 4! = 33 which ends with 3.

We need to find, $3^{1729}$ which is nothing but power cycles of 3 which will end with 3.