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What is the units digit of 3^1001 × 7^1002 × 13^1003 ?
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3^1001 × 7^1002 × 13^1003 = (3*7*13)^1001 * 7*(13^2) = 273^1001 * 1183
Regarding 273^n, the last digit is periodical with the period of 4, so 273^1001 has the last digit 3.
Therefore, the given number has the same last digit as 3*1183, i.e. 9.
Solved.
The answer is 9.
