Solution - Long multiplication

Start by multiplying the ones digit (2) of the multiplier 122 by each digit of the multiplicand 61, 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:
2×6=12

Write 2 in the tens place.

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

122 is the first partial product.

Proceed by multiplying the tens digit (2) of the multiplier (122) by each digit of the multiplicand (61), from right to left.

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

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

Write 2 in the tens place.

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

Write 2 in the hundreds place.

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

1,220 is the second partial product.

Proceed by multiplying the hundreds digit (1) of the multiplier (122) by each digit of the multiplicand (61), 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:
1×6=6

Write 6 in the thousands place.

6,100 is the third partial product.

122+1220+6100=7442 long addition steps can be seen here

The solution is: 7,442