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Smallwonder
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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 12 Sep 2013, 05:19
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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75% (hard)

Question Stats:

55% (02:19) correct 45% (02:21) wrong based on 511 sessions

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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
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D
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Vetrik
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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 09 Nov 2014, 10:22
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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) 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'

Hence Answer D
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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 12 Sep 2013, 05:30
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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.
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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 12 Sep 2013, 05:56
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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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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 04 Aug 2014, 11:17
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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.

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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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 24 Aug 2014, 02:19
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Does someone have an answer here? i still don't get how to solve it.
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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 26 Feb 2016, 19:43
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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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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 26 Feb 2016, 19:57
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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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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 17 Mar 2016, 07:00
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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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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 07 Nov 2016, 23:43
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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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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 26 Jun 2020, 15:12
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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.

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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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 08 Feb 2024, 07:53
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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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Find unit digit of N when N = 63^(1! + 2! + ... + 63!) + 18^(1! + 2! + 27 Feb 2025, 09:19
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