1. **State the problem:** Multiply the whole numbers 253 and 36. 2. **Write the multiplication setup:** $$253 \times 36$$ 3. **Multiply the ones digit of the second number (6) by each digit of the first number (253):** $$6 \times 3 = 18$$ (write 8, carry 1) $$6 \times 5 = 30$$ plus carry 1 equals 31 (write 1, carry 3) $$6 \times 2 = 12$$ plus carry 3 equals 15 (write 15) So, the first partial product is: $$1518$$ 4. **Multiply the tens digit of the second number (3, which is actually 30) by each digit of the first number (253):** $$3 \times 3 = 9$$ (write 9) $$3 \times 5 = 15$$ (write 5, carry 1) $$3 \times 2 = 6$$ plus carry 1 equals 7 (write 7) Since this is the tens place, add a zero at the end: $$7590$$ 5. **Add the two partial products:** $$\begin{array}{r} \phantom{0}1518 \\ +7590 \\ \hline 9108 \end{array}$$ 6. **Final answer:** $$253 \times 36 = 9108$$ This means when you multiply 253 by 36, the product is 9108.