What is the sum of the first 10 natural numbers?
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The sum of the first n natural numbers is n(n + 1)/2.
For n = 10, the sum is 10 × 11 ÷ 2 = 55.
Practice NET Engineering questions with answers and explanations.
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The sum of the first n natural numbers is n(n + 1)/2.
For n = 10, the sum is 10 × 11 ÷ 2 = 55.
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The sum of the first n odd numbers equals n².
For n = 8, the sum is 8² = 64.
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The least common multiple of 3 and 4 is 12.
A number divisible by both must therefore be divisible by 12.
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The units digits of powers of 7 cycle as 7, 9, 3, 1.
The fourth power therefore ends in 1.
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The greatest common divisor of 18 and 24 is 6.
Dividing both terms by 6 gives 3/4.
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Convert 2/3 to sixths: 2/3 = 4/6.
Then 4/6 + 1/6 = 5/6.
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Convert 1/4 to 2/8.
Then 7/8 - 2/8 = 5/8.
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Multiply numerators and denominators, then simplify.
(3 × 10)/(5 × 9) = 30/45 = 2/3.
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Dividing by a fraction means multiplying by its reciprocal.
Thus, 4/7 × 3/2 = 12/14 = 6/7.
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The decimal 0.375 equals 375/1000.
Dividing numerator and denominator by 125 gives 3/8.
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Divide 7 by 20 or convert the denominator to 100.
Since 7/20 = 35/100, the decimal is 0.35.
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Twenty-five percent is one quarter.
One quarter of 240 is 240 ÷ 4 = 60.