Consecutive is unpredictable (Part 2)

Let a b c \large{\color{#D61F06}a\color{#3D99F6}b\color{#69047E}c} be the last three digits of 1 2 + 2 3 + 3 4 + 4 5 + 5 6 + . . . + 99 8 999 + 99 9 1000 \large{1^{2}+2^{3}+3^{4}+4^{5}+5^{6}+...+998^{999}+999^{1000}} .

Find a b c \large{\color{#D61F06}a\color{#3D99F6}b\color{#69047E}c} m o d mod 69 69 .

Details:

The exponents are in a consecutive order of the positive integers: 2 , 3 , 4 , . . . , 998 , 999 , 1000 2, 3, 4, ..., 998, 999, 1000

a b c \large{\color{#D61F06}a\color{#3D99F6}b\color{#69047E}c} is three different digits, like 346 346 , not algebraic expression like a × b × c \large{\color{#D61F06}a \times\color{#3D99F6}b\times\color{#69047E}c} ( 3 × 4 × 6 ) (3\times4\times6)

More challenging problems? Here!


The answer is 10.

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1 solution

effortless :) (Frequent use of modulo prevents the variable to balloon in value)

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