Solution - Long multiplication

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

Proceed by multiplying the hundreds digit (1) of the multiplier (1,100) by each digit of the multiplicand (2,500), 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:
1×0=0

Write 0 in the thousands place.

Multiply the hundreds digit (1) of the multiplicator by the number in the hundreds place value:
1×5=5

Write 5 in the ten thousands place.

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

Write 2 in the hundred thousands place.

250,000 is the first partial product.

Proceed by multiplying the thousands digit (1) of the multiplier (1,100) by each digit of the multiplicand (2,500), 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:
1×0=0

Write 0 in the thousands place.

Multiply the thousands digit (1) of the multiplicator by the number in the tens place value:
1×0=0

Write 0 in the ten thousands place.

Multiply the thousands digit (1) of the multiplicator by the number in the hundreds place value:
1×5=5

Write 5 in the hundred thousands place.

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

Write 2 in the millions place.

2,500,000 is the second partial product.

250000+2500000=2750000 long addition steps can be seen here

The solution is: 2,750,000