2006 AMC 10B Problem 11 — Solution | LIVE by Po-Shen Loh

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2006 AMC 10B Problem 11

Below is the professionally curated solution for Problem 11 of the 2006 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:factorial modular arithmetic digits

Difficulty rating: 1280

11.

What is the tens digit in the sum $7! + 8! + 9! + \cdots + 2006!\,?$

$1$

$3$

$4$

$6$

$9$

Solution:

For $n\ge 10,$ $n!$ is divisible by $100,$ so it does not affect the last two digits.

The tens digit comes from $7!+8!+9!=5040+40320+362880=408240,$ whose tens digit is $4.$

Thus, the correct answer is C.

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Problem 11 in Other Years

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