What is longmult 32*32 ?

The solution to longmult 32*32 is 1024

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longmult 32*32

long multiplication

32 \cdot 32

1024

Solution steps

32 \cdot 32

Line up the numbers

\times 3322

Multiply the top number by the bottom number one digit at a time starting with the ones digit (from right to left)

Multiply the top number by the bolded digit of the bottom number

\times 3322

Mutliply the bold numbers:

2 \cdot 2 = 4

\frac{\times 3 3 2 2}{4}

Mutliply the bold numbers:

3 \cdot 2 = 6

\frac{\times 3 3 2 2}{6 4}

Multiply the top number by the bolded digit of the bottom number

\frac{\times 3 3 2 2}{6 4}

Mutliply the bold numbers:

2 \cdot 3 = 6

\frac{\times 3 3 2 2}{6 6 4}

Mutliply the bold numbers:

3 \cdot 3 = 9

\frac{\times 3 3 2 2}{9 6 6 4}

Add the rows to get the answer. For simplicity, fill in trailing zeros

\frac{\times 3 3 2 2}{0 9 6 6 4 0}

64 + 960 = 1024

+ 096640

Add the digits of the bolded column:

4 + 0 = 4

\frac{+ 0 9 6 6 4 0}{+ 4}

Add the digits of the bolded column:

6 + 6 = 12

\frac{+ 0 9 6 6 4 0}{+ 4}

Carry

1

to the column on the left and write

2

in the bolded column

\frac{+ 1 0 9 6 6 4 0}{+ 2 4}

Add the digits of the bolded column:

1 + 0 + 9 = 10

\frac{+ 1 0 9 6 6 4 0}{+ 2 4}

Carry

1

to the column on the left and write

0

in the bolded column

\frac{+ 1 1 0 9 6 6 4 0}{+ 0 2 4}

Add the carried digit,

1

, to the result

\frac{+ 1 1 0 9 6 6 4 0}{1 0 2 4}

= 1024

= 1024

2^{2} + 1

- 2 (1)^{2}

(9 \cdot 8)^{4}

6 \cdot 6 \cdot 6

4 (- 2)^{2} + 8 (- 2) + 3 (- 2) + 6