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.