Solution - Long multiplication

Start by multiplying the ones digit (5) of the multiplier 25 by each digit of the multiplicand 25, from right to left.

Multiply the ones digit (5) of the multiplicator by the number in the ones place value:
5×5=25

Write 5 in the ones place.

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

Multiply the ones digit (5) of the multiplicator by the number in the tens place value and add the carried number (2):
5×2+2=12

Write 2 in the tens place.

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

125 is the first partial product.

Proceed by multiplying the tens digit (2) of the multiplier (25) by each digit of the multiplicand (25), 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:
2×5=10

Write 0 in the tens place.

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

Multiply the tens digit (2) of the multiplicator by the number in the tens place value and add the carried number (1):
2×2+1=5

Write 5 in the hundreds place.

500 is the second partial product.

125+500=625 long addition steps can be seen here

The solution is: 625