Olympiad Maths Prep

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Problem 387

AMC 12 late, AIME early
Number theory Difficulty 4.8 Find the answer

18) The sum of all two-digit numbers with distinct digits is:
(A) 3840
(B) 3960,
(C) 4140,
(D) 4260,
(E) 4410.

Official solution

18. The answer is (E)(\mathbf{E}).

The sum of all two-digit numbers is

10+11++98+99=(1+2++98+99)(1+2++8+9)=99100245=4510910+11+\cdots+98+99=(1+2+\cdots+98+99)-(1+2+\cdots+8+9)=\frac{99 \cdot 100}{2}-45=45 \cdot 109.

From this sum, I need to subtract the sum of the numbers formed by two identical digits:

11+22++88+99=11(1+2++8+9)=1145. 11+22+\cdots+88+99=11(1+2+\cdots+8+9)=11 \cdot 45 .

The result is therefore 451094511=4598=441045 \cdot 109-45 \cdot 11=45 \cdot 98=4410.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.