Question

The unit place of 94^(1!+2!+3!+⋯..+94!)×76^(1!+2!+3!+⋯…+76!)×103^(1!+2!+3!+⋯.+103!)​

Answer 1

Answer:

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Step-by-step explanation:

We have to evaluate unit's digit of the sum 1!+2!+3!+...+9!

We have,

1!=1

2!=2

3!=6

4!=24

5!=120

6!=6(5!)=720

7!=7×6(5!)= a multiple of 10.

Clearly, factorial of every number after 4 will be a multiple of 10. So, factorial value of every number after 4 will have 0 at its unit's place.

Unit digit of required sum 1+2+6+4+0+... is 3.