Solution - Long multiplication

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

Proceed by multiplying the hundreds digit (3) of the multiplier (300) by each digit of the multiplicand (25,000), from right to left.

Because digit (3) is in hundreds place, we shift partial result by 2 place(s) by placing 2 zero(s).

Multiply the hundreds digit (3) of the multiplicator by the number in the ones place value:
3×0=0

Write 0 in the hundreds place.

Multiply the hundreds digit (3) of the multiplicator by the number in the tens place value:
3×0=0

Write 0 in the thousands place.

Multiply the hundreds digit (3) of the multiplicator by the number in the hundreds place value:
3×0=0

Write 0 in the ten thousands place.

Multiply the hundreds digit (3) of the multiplicator by the number in the thousands place value:
3×5=15

Write 5 in the hundred thousands place.

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

Multiply the hundreds digit (3) of the multiplicator by the number in the ten thousands place value and add the carried number (1):
3×2+1=7

Write 7 in the millions place.

7,500,000 is the first partial product.

7500000=7500000 long addition steps can be seen here

The solution is: 7,500,000