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hareesh860
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starting from 5! we get trailing zeros at the end, hence we need to sum up only 1!+2!+3!+4! which is 33 with 3 as a unit's digit.hareesh860 wrote:Let x = 1! + 2! + 3! + 4! + ... + 100!. Unit's digit of X^X^X^X........infinity
(a) 3
(b) 1
(c) or 3 depending upon the number of times x appears in the power.
(d) Can't be determined.
(e) 4
The required is X^X^X^X........infinity or abdcef .....3^abdcef .....3^abdcef .....3^abdcef .....3^ ... infinity
Since the cycle of 3 is 1-5 coefficients it will be not difficult to determine the last digit (see below)
3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
because the numbers ending with a unit's digit 3 are all divisible by the cycles of 3 with only two options of 3 and 7 as unit's digit depending on the number of tens in the numbers. If the ten's number of X is odd, then the unit's digit is 3. If the ten's number of X is even, then the unit's digit will be 7.
when the unit's digit is 7 the next number X raised to the power of X with the unit's digit of 3 will return only 3 or 7 as the unit's digit, depending again on the number of odd or even tens'.
only option c suits here, does this option contain 7 or 3? the question is addressed to @hareesh860. If so you need to update the post.
