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Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+.
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09 Nov 2014, 10:22
7
2
Smallwonder wrote:
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!
A) 2 B) 437 D) 8 E) 0
The cyclicity of 3, 7 & 8 are 4 . Hence we need to divide the power by 4 and the reminder should be taken as the new power and from there unit digit of each of three terms needs to be found out. By adding the unit digit of the three terms we get the unit digit of the complete expression.
But how do find the reminder of the power divided by 4?
1 ! = 1 ; 2 ! = 2 ; 3 ! = 6 ...... 4 ! is 1*2*3*4 hence divisible by 4...similarly all the factorials above 4! will also be divisible by 4.
The power of 63 is 1+2+6+ 4 (1*2*3 + 1* 2*3*5 +.....) = 9 + 4(X) / 4 -> reminder is 1. Similarly the new powers of 18 & 37 are also 1.
Hence the unit digits are 63^ 1 + 18 ^ 1 + 37 ^ 1 -> 3+ 8+ 7 -> 18 ...so the last digit of is '8'
All the powers given(1!+2!+....), are multiples of 4. It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.
Thus, the given problem boils down to \(3^4+8^4+7^4 = 1+6+1 = 8\)
After 4! + any terms above will be divisible by 4 and hence the power term becomes 9 + (Term which is divisible by 4). Therefore the remainder will always be 1 in all the 3 cases.
Now considering the power cycle for 3,8 and 7 we get 3+8+7 = 18. Hence unit digit is 8.
All the powers given(1!+2!+....), are multiples of 4. It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.
Thus, the given problem boils down to \(3^4+8^4+7^4 = 1+6+1 = 8\)
Thus, the units digit is 8.
D.
Hi , Can u please explain me how did we that last two digits of the total sum will be 00, for each of them?
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+.
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26 Feb 2016, 19:43
I don't see how it can be 8..and this is how I tried to check... power of 3 - units digit 1 - 3 2 - 9 3 - 7 4 - 1 so for every power divisible by 4, the units digit is 1.
now 8: 1 - 8 2 - 4 3 - 2 4 - 6 5 - 8 we can ID that for every power divisible by 4, the units digit is 6.
now, all the possible units digit for first one: 3,9,7,1 all possible units digit for second one: 8,4,2,6 3+8=11 - no 3+4=7 - no 3+2=5-no 3+6=9-no
9+8=17 - no 9+4=13 - no 9+2=11 - no 9+6=15 - no
7+8=15 - no 7+4=11 - no 7+2=9 - no 7+6 = 13 - no
1+8=9 - no 1+4=5 - no 1+2=3 - no 1+6 = 7 - no
all the possible answers - neither of the options is present. hm?
oh man..now I see where i made the mistake..I did not see 37^....
Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+.
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26 Feb 2016, 19:57
mvictor wrote:
I don't see how it can be 8..and this is how I tried to check... power of 3 - units digit 1 - 3 2 - 9 3 - 7 4 - 1 so for every power divisible by 4, the units digit is 1.
now 8: 1 - 8 2 - 4 3 - 2 4 - 6 5 - 8 we can ID that for every power divisible by 4, the units digit is 6.
now, all the possible units digit for first one: 3,9,7,1 all possible units digit for second one: 8,4,2,6 3+8=11 - no 3+4=7 - no 3+2=5-no 3+6=9-no
9+8=17 - no 9+4=13 - no 9+2=11 - no 9+6=15 - no
7+8=15 - no 7+4=11 - no 7+2=9 - no 7+6 = 13 - no
1+8=9 - no 1+4=5 - no 1+2=3 - no 1+6 = 7 - no
all the possible answers - neither of the options is present. hm?
oh man..now I see where i made the mistake..I did not see 37^....
You are not taking 37^abc... into account while taking the sum. Your calculations are only taking 63^1!+2!+...+63! + 18^1!+2!....
All the powers given(1!+2!+....), are multiples of 4.It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.
Thus, the given problem boils down to \(3^4+8^4+7^4 = 1+6+1 = 8\)
Thus, the units digit is 8.
D.
I don't think the last digit of all powers would be 00, it would be 13 which will leave a remainder of 1 when divided by 4 and then the cyclicity rule can be applied.
We will get it at 3 + 7 + 8 = 18 and last digit would be 8.
Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+.
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24 Dec 2017, 13:13
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