Solution - Long multiplication

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

Proceed by multiplying the tens digit (6) of the multiplier (160) by each digit of the multiplicand (320), from right to left.

Because digit (6) is in tens place, we shift partial result by 1 place(s) by placing 1 zero(s).

Multiply the tens digit (6) of the multiplicator by the number in the ones place value:
6×0=0

Write 0 in the tens place.

Multiply the tens digit (6) of the multiplicator by the number in the tens place value:
6×2=12

Write 2 in the hundreds place.

Because the result is greater than 9, carry the 1 to the thousands place.

Multiply the tens digit (6) of the multiplicator by the number in the hundreds place value and add the carried number (1):
6×3+1=19

Write 9 in the thousands place.

Because the result is greater than 9, carry the 1 to the ten thousands place.

19,200 is the first partial product.

Proceed by multiplying the hundreds digit (1) of the multiplier (160) by each digit of the multiplicand (320), 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:
1×0=0

Write 0 in the hundreds place.

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

Write 2 in the thousands place.

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

Write 3 in the ten thousands place.

32,000 is the second partial product.

19200+32000=51200 long addition steps can be seen here

The solution is: 51,200