Solution - Long multiplication

Start by multiplying the ones digit (4) of the multiplier 1,224 by each digit of the multiplicand 7, from right to left.

Multiply the ones digit (4) of the multiplicator by the number in the ones place value:
4×7=28

Write 8 in the ones place.

Because the result is greater than 9, carry the 2 to the tens place.

28 is the first partial product.

Proceed by multiplying the tens digit (2) of the multiplier (1,224) by each digit of the multiplicand (7), 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:
2×7=14

Write 4 in the tens place.

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

140 is the second partial product.

Proceed by multiplying the hundreds digit (2) of the multiplier (1,224) by each digit of the multiplicand (7), from right to left.

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

Multiply the hundreds digit (2) of the multiplicator by the number in the ones place value:
2×7=14

Write 4 in the hundreds place.

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

1,400 is the third partial product.

Proceed by multiplying the thousands digit (1) of the multiplier (1,224) by each digit of the multiplicand (7), from right to left.

Because digit (1) is in thousands place, we shift partial result by 3 place(s) by placing 3 zero(s).

Multiply the thousands digit (1) of the multiplicator by the number in the ones place value:
1×7=7

Write 7 in the thousands place.

7,000 is the fourth partial product.

28+140+1400+7000=8568 long addition steps can be seen here

The solution is: 8,568