What is 10^9-1\div 999 ?

The solution to 10^9-1\div 999 is 998999999999/999

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10^9-1\div 999

1 0^{9} - 1 \div 999

\frac{998999999999}{999}

Solution steps

1 0^{9} - 1 \div 999

Follow the PEMDAS order of operations

Calculate exponents

1 0^{9} : 1000000000

1 0^{9}

1 0^{9} = 1000000000

= 1000000000

= 1000000000 - 1 \div 999

Multiply and divide (left to right)

1 \div 999 : \frac{1}{999}

1 \div 999

1 \div 999 = \frac{1}{999}

= \frac{1}{999}

= 1000000000 - \frac{1}{999}

Add and subtract (left to right)

1000000000 - \frac{1}{999} : \frac{998999999999}{999}

1000000000 - \frac{1}{999}

1000000000 - \frac{1}{999} = \frac{998999999999}{999}

1000000000 - \frac{1}{999}

Convert element to fraction:

1000000000 = \frac{1000000000 \cdot 999}{999}

= \frac{1000000000 \cdot 999}{999} - \frac{1}{999}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1000000000 \cdot 999 - 1}{999}

1000000000 \cdot 999 - 1 = 998999999999

1000000000 \cdot 999 - 1

Multiply the numbers:

1000000000 \cdot 999 = 999000000000

= 999000000000 - 1

Subtract the numbers:

999000000000 - 1 = 998999999999

= 998999999999

= \frac{998999999999}{999}

= \frac{998999999999}{999}

= \frac{998999999999}{999}

(- 3) \cdot [(- 2) + (- 5)]

\frac{4 ^{2} ( 22 - 18 )}{3 ( - 7 ) - ( - 4 ) ( - 2 )}

\frac{1}{12} \times 5 \times 8^{3}

37 + 8 - 29

4^{3} \div 2 - 2^{3} \times 6 + 8 (- 1)^{10}