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Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0

Considering 1! + 2! + 3! + 4! + 5! + ..... = 1 + 2 + 6 + (24 + 120 + 720 + XXX0 ... )

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.
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Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0

3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.

For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243

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?
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Does someone have an answer here? i still don't get how to solve it.
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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^....
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mvictor
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!....

Refer to find-unit-digit-of-n-when-n-159623.html#p1266283 for the solution.
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Superb Question...
Here the key is to write 1!+2+....as 9+4*p for some integer p
hence UD => UD of 3+7+8=> 8
hence D
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mau5
Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0

3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.

For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243

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.
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Smallwonder
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!

A) 2
B) 4
C) 6
D) 8
E) 0

3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.

For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243

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,

What if we solve the question with another approach.?

What I have done is that I added 1! up to 4! which gives us 33 because after 4! the unit's digit will be 0 and so, if we add the numbers in each case the unit's digit will be 3.

Thus, it would look like 63^ 3 + 18^3 + 37^3 ( I have just written unit digit)
Hence, when we will solve the units digit will appear as 7+2+3 = 12 which gives us 2 as the answer. ( 3^3= 27, 8^3=512 , 7^3= 343)

Could you please tell me what is wrong with my approach?
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1!+2!+3!+4!...=4k+1

63^1!+2!+...+63! 
63^(4k+1): last digit (3 has a cycle of 3,9,7,1), =3


18^1!+2!+...+18! 
18^(4k+1): last digit (8 has a cycle of 8,4,2,6), =8

37^1!+2!+...37!
37^(4k+1): last digit (7 has a cycle of 7,9,3,1), =7

Unit's digit=3+8+7=18=8
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