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OpenStudy (anonymous):
help me on my question please!
ganeshie8 (ganeshie8):
use this :- starting from 5, every integer factorial has 0 in its units place
ganeshie8 (ganeshie8):
All you have to do is to find last digit of, 1! + 2! + 3! + 4!
OpenStudy (anonymous):
1
2
6
24
OpenStudy (anonymous):
total of 1+2+6+24=33
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ganeshie8 (ganeshie8):
Yes. just take the last digit of that
ganeshie8 (ganeshie8):
cuz for numbers greater than 4, last digit is 0. so they wont affect the last digit.
OpenStudy (anonymous):
33
ganeshie8 (ganeshie8):
last digit is just 3
OpenStudy (anonymous):
okay
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OpenStudy (anonymous):
what after that
ganeshie8 (ganeshie8):
Nothing. we're done. last digit is 3.
OpenStudy (anonymous):
okay
OpenStudy (anonymous):
@ganeshie8
thank u
OpenStudy (anonymous):
@Loser66
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OpenStudy (loser66):
you want to know the logic?
ganeshie8 (ganeshie8):
yw!
OpenStudy (loser66):
\[1! =1\\2! = 1!*2 =1*2=2\\3!= 2!*3=2*3=6\\4!=3!*4=6*4=24\\5!=4!*5=24*5 = 120\\6!=5!*6= 720\]
pay attention at 5! since the last digit of 5! is 0 so, the next term will have the last digit is 0
therefore, your last digit is counted from the first four terms.