Find the unit place of 2^1!+2!+3!+ .....+10! !

Question

Find the unit place of

2^(1!+2!+3!+...+10!) + 3^(1!+2!+3!+...+10!) + 4^(1!+2!+3!+...+10!) + 5^(1!+2!+3!+...+10!) + ... + 9^(1!+2!+3!+...+10!) ?

A. 4
B. 5
C. 0
D. 6
E. 9

sujoykrdatta · Answer

The exponent for each term is 1! + 2! + 3! + ... + 10! 
From 4! onward, each term is a multiple of 4

1! + 2! +3! = 1 + 2 + 6 = 9 = 4 * 2 + 1 = 'A multiple of 4' + 1

Thus, (1! + 2! + 3! + 4! + ... + 10!) is 'a multiple of 4' + 1

=> 1! + 2! + 3! + 4! + ... + 10! = 4k + 1 (where k is some positive integer)

Thus, we have: 2^(4k+1) + 3^(4k+1) + 4^(4k+1) + ... + 9^(4k+1)

Each of the digits (in the base) 2, 3, 4, 5, 6, 7, 8 and 9 have a cyclicity of 4 places (4 and 9 have a cyclicity of 2 places, but for all practical purposes can be thought of having a cyclicity of 4 places)

Thus, we have:

Units digit of 2^(4k+1) + 3^(4k+1) + 4^(4k+1) + ... + 9^(4k+1)

= Units digit of 2^1 + 3^1 + 4^1 + ... + 9^1

= Units digit of 2 + 3 + 4 + ... + 9

= Units digit of 44

= 4

Answer A

yutaro0121 · Answer

This is not good format.
The correct question is

Bunuel · Answer

Fixed the issue. Thank you!

impatientdude · Answer

Convert this to 2^y + 3^y + 4^y + 5^y

Compute unit digit cycle:  
2's: 2-4-8-6
3's: 3-9-7-1
4's: 4-6-4-6
5's: 5-5-5-5

now 1!+2!+...10! would have 1 odd + 9 evens (everything after 2! is even) => odd

so we only need to consider odd positions in unit digit cycle i.e.:
2+3+4+5 = 1 4 
or 8+7+4+5 = 2 4

Thus ans: 4

KarishmaB · Answer

Note that 2! onwards every factorial is even. Only 1! = 1. So sum of first ten factorials will be 1 + Even = Odd number. It will be of the form (4n + 1) or (4n + 3), but we do not know which. The units digit of the following are different for the 2 forms:

Cyclicity of 2 is 2, 4, 8, 6
Cyclicity of 3 is 3, 9, 7, 1
Cyclicity of 7 is 7, 9, 3, 1
Cyclicity of 8 is 8, 4, 2, 6

For others, the units digit are the same i.e. 4^1 and 4^3 have the same units digit of 4. Similarly for 5, 6, 9.

If the exponent is of the form (4n + 1), then the sum of all units digits will be 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 44
If the exponent is of the form (4n + 3), then the sum of all units digits will be 8 + 7 + 4 + 5 + 6 + 3 + 2 + 9 = 44

So units digit will always be 4.

Answer (A)

Find the unit place of 2^1!+2!+3!+ .....+10! !

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Find the unit place of 2^1!+2!+3!+ .....+10! !

Prep Toolkit

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