Question

what is the remainder when 1! +2! +3! +....+2006! is divided by 21 ? (Please tell with explaination)

Answer 1

Answer:

14

Step-by-step explanation:

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Answer 2

Answer:

what is the remainder when 1! +2! +3! +....+2006! is divided by 21 ?

Note that 3 * 7 = 21 and gcd(3, 7) = 1; so we can break this problem into smaller pieces.

i) Remainder upon division by 3:

Note that 3! is divisible by 3 ==> k! is divisible by 3 for all k = 3, 4, ...

So 1!+2!+3!+4! +...+2006! has remainder from 1! + 2! = 3, which is 0.

ii) Remainder upon division by 7:

Note that 7! is divisible by 7 ==> k! is divisible by 7 for all k = 7, 8, ...

So 1!+2!+3!+4! +...+2006! has remainder from 1! + 2! + 3! + 4! + 5! + 6! = 873, which is 5.

By (i) and (ii):

The integer between 0 and 21 (not including 21) which leaves remainder 0 and 5 upon division by 3 and 7, respectively, is readily checked to be 12.

Hence, the remainder of (1! + 2! + ... + 2006!)/21 equals 12.

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P.S.:

If you know modular arithmetic, you can clean up the arguments above very nicely:

Working mod 3,

1!+2!+3!+4! +...+2006! = 1! + 2! + 0 = 3 = 0 (mod 3).

Working mod 7:

1!+2!+3!+4! +...+2006! = 1! + 2! + ... + 6! + 0 = 873 = 5 (mod 7).

Finally, by Chinese Remainder Theorem or otherwise,

n = 0 (mod 3) and n = 5 (mod 7) ==> n = 12 (mod 21).

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