Solution - Long multiplication

Start by multiplying the ones digit (2) of the multiplier 102 by each digit of the multiplicand 31, from right to left.

Multiply the ones digit (2) of the multiplicator by the number in the ones place value:
2×1=2

Write 2 in the ones place.

Multiply the ones digit (2) of the multiplicator by the number in the tens place value:
2×3=6

Write 6 in the tens place.

62 is the first partial product.

Because the tens digit of the multiplicator equals 0, skip to the next digit.

Proceed by multiplying the hundreds digit (1) of the multiplier (102) by each digit of the multiplicand (31), from right to left.

Because digit (1) is in hundreds place, we shift partial result by 2 place(s) by placing 2 zero(s).

Multiply the hundreds digit (1) of the multiplicator by the number in the ones place value:
1×1=1

Write 1 in the hundreds place.

Multiply the hundreds digit (1) of the multiplicator by the number in the tens place value:
1×3=3

Write 3 in the thousands place.

3,100 is the second partial product.

62+3100=3162 long addition steps can be seen here

The solution is: 3,162