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Quiz about Whats the Sum
Quiz about Whats the Sum

What's the Sum? Trivia Quiz


You know 1 + 1 = 2, but how about these 10 questions? Enjoy!

A multiple-choice quiz by Matthew_07. Estimated time: 6 mins.
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Author
Matthew_07
Time
6 mins
Type
Multiple Choice
Quiz #
301,458
Updated
Dec 03 21
# Qns
10
Difficulty
Tough
Avg Score
6 / 10
Plays
3275
Awards
Top 20% Quiz
Last 3 plays: HumblePie7 (4/10), Hamnah9 (4/10), shvdotr (8/10).
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Question 1 of 10
1. Which Greek letter is used to denote summation in mathematics? Hint


Question 2 of 10
2. Pick any number from 2 to 9. Now, multiply that number by 9. You will get a 2-digit number. Sum up the 2 individual digits and type in your answer.

Answer: (A 1-digit number)
Question 3 of 10
3. To determine whether a number can be divided completely by 3 without any remainder, we can sum up their individual digits. If the sum is divisible by 3, then the number itself can also be divisible by 3. Which of the following numbers is NOT divisible by 3? Hint


Question 4 of 10
4. The sum of the three most beautiful and intriguing mathematical constants, pi, e and Phi is closest to which integer? Hint


Question 5 of 10
5. Evaluate the sum of all the positive integers between 1 and 100 (inclusive) that are divisible by 10. Hint


Question 6 of 10
6. The sum of the first 10 positive integers, 1 + 2 + ... + 9 + 10 is 55. What's the trick here? Well, notice that 1 + 2 + ... + 9 + 10 = 1 + 10 + 2 + 9 +... = (1 + 10) + (2 + 9) + ... = 11 + 11 + ... = 5 (11) = 55. Now, your turn. What is the sum of the first 1000 positive integers?

Answer: (A 6-digit number)
Question 7 of 10
7. Given that the sum of the first 100 odd integers, namely 1 + 3 + ... + 197 + 199 = 10000. Now can you tell me what is the sum of the first 100 even integers, namely 2 + 4 + ... + 198 + 200? Hint


Question 8 of 10
8. Is the sum of 1 + 2 - 3 + 4 - ... - 97 + 98 - 99 + 100 the same as the sum of 1 - 2 + 3 - 4 ... + 97 - 98 + 99 - 100?


Question 9 of 10
9. By using the partial fraction technique, the fraction 1/[(n)(n+1)] can be rewritten as 1/n - 1/(n+1). For example, 1/(2x3) = 1/6 is equivalent to 1/2 - 1/3. What is the sum of 1/2 + 1/6 + 1/12 + ... + 1/9702 + 1/9900 = 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(98x99) + 1/(99x100)? Hint


Question 10 of 10
10. A = the number of sides of a pentagon; B = the third odd positive integer; C = the fifth Fibonacci number (the sequence starts with 1, 1, 2...); D = the number of diagonals of a pentagon; E = the third prime number.

What is the sum of A, B, C, D and E? In other words, compute A + B + C + D + E.

Answer: (A 2-digit number)

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quiz
Quiz Answer Key and Fun Facts
1. Which Greek letter is used to denote summation in mathematics?

Answer: Sigma

The upper case sigma is the notation for summation or addition operation. The lower case sigma is used to denote the standard deviation of a set of data in statistics.
2. Pick any number from 2 to 9. Now, multiply that number by 9. You will get a 2-digit number. Sum up the 2 individual digits and type in your answer.

Answer: 9

No matter what number you pick, the final answer would be 9.

Let's look at a few examples. Let say you pick number 2. 2 x 9 = 18. 1 + 8 = 9.

If you pick 5, 5 x 9 = 45. 4 + 5 = 9.

And if you pick 9, 9 x 9 = 81. 8 + 1 = 9.

The multiples of 9 are 18, 27, 36, 45, 54, 63, 72 and 81. No matter what multiple you arrive at, the sum of its individual digit is 9.
3. To determine whether a number can be divided completely by 3 without any remainder, we can sum up their individual digits. If the sum is divisible by 3, then the number itself can also be divisible by 3. Which of the following numbers is NOT divisible by 3?

Answer: 767676767

Let's look at the number 767676767. The sum of the individual digits is 59, which is not divisible by 3.

Now, the second number, 123321333. The sum of the individual digits is 21, which is divisible by 3. So the number 12332133 itself can also be divisible by 3.

Also, the sums of the individual digits of 696222333 and 998877663 are 36 and 63 respectively. 36 / 3 = 12. Besides, 63 / 3 = 21. Therefore, both 696222333 and 998877663 are divisible by 3.
4. The sum of the three most beautiful and intriguing mathematical constants, pi, e and Phi is closest to which integer?

Answer: 7

Pi (3.142..), e (2.718...) and Phi (1.618...) are all irrational numbers. Their sum is 7.478..., which is close to 7.
5. Evaluate the sum of all the positive integers between 1 and 100 (inclusive) that are divisible by 10.

Answer: 550

In other words, the question is asking for the sum of 10 + 20 + 30 +...+ 80 + 90 + 100 = 10 + 100 + 20 + 90 + 30 + 80 + ... = (10 + 100) + (20 + 90) + (30 + 80) + ... = 110 + 110 + 110 + ... = 5(110) = 550
6. The sum of the first 10 positive integers, 1 + 2 + ... + 9 + 10 is 55. What's the trick here? Well, notice that 1 + 2 + ... + 9 + 10 = 1 + 10 + 2 + 9 +... = (1 + 10) + (2 + 9) + ... = 11 + 11 + ... = 5 (11) = 55. Now, your turn. What is the sum of the first 1000 positive integers?

Answer: 500500

By applying the same technique, 1 + 2 + ... + 999 + 1000 = 1 + 1000 + 2 + 999 + 3 + 998 + ... = (1 + 1000) + (2 + 999) + (3 + 998) + ... = (1001) + (1001) + (1001) + ... = 500 (1001) = 500500

Keep in mind that there are all together 1000 / 2 = 500 pairs of doublet here. The sum of each doublet is 1001. So 500 x 1001 = 500500
7. Given that the sum of the first 100 odd integers, namely 1 + 3 + ... + 197 + 199 = 10000. Now can you tell me what is the sum of the first 100 even integers, namely 2 + 4 + ... + 198 + 200?

Answer: 10100

Notice that there are all together 100 terms in our summation of the first 100 even numbers. Also, there are 100 terms in the summation of the first 100 odd numbers.

2 + 4 + ... + 198 + 200 = (1 + 1) + (3 + 1) + ... + (197 + 1) + (199 + 1) = (1 + 3 + ... + 197 + 199) + (1 + 1 + ... + 1 + 1) = 10000 + 100 (1) = 10000 + 50 = 10100

The trick here is that we can make use of the first equation given in the question. Observe that 2 is one more than 1, 4 is one more than 3 and so on. After some rearranging and algebraic manipulation, we only need to evaluate the sum of 10000 and 100, which gives us the answer of 10100.
8. Is the sum of 1 + 2 - 3 + 4 - ... - 97 + 98 - 99 + 100 the same as the sum of 1 - 2 + 3 - 4 ... + 97 - 98 + 99 - 100?

Answer: No

1 + 2 - 3 + 4 ... -97 + 98 - 99 + 100 = 1 + (2 - 3) + (4 - 5) ... + (96 - 97) + (98 - 99) + 100 = 1+ (-1) + (-1) ... + (-1) + (-1) + 100 = 1 + 49 (-1) + 100 = 1 - 49 + 100 = 52

On the other hand, 1 - 2 + 3 - 4 ... + 97 - 98 + 99 - 100 = (1 - 2) + (3 - 4) ... + (97 - 98) + (99 - 100) = (-1) + (-1) ... + (-1) + (-1) = 50 (-1) = -50

So the answer is no. We can solve this problem by a simpler approach, namely by breaking down the summation into several smaller parts. By observing the pattern, we will found out that there are many repeating -1's. We can group them together and use the multiplication operation which is much simpler compared to the tedious repetitive addition operation.
9. By using the partial fraction technique, the fraction 1/[(n)(n+1)] can be rewritten as 1/n - 1/(n+1). For example, 1/(2x3) = 1/6 is equivalent to 1/2 - 1/3. What is the sum of 1/2 + 1/6 + 1/12 + ... + 1/9702 + 1/9900 = 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(98x99) + 1/(99x100)?

Answer: 0.99

Instead of adding up 1/2 + 1/6 + 1/12 + ... + 1/9702 + 1/9900 one by one, we can use the formula 1/[(n)(n+1)] = 1/n - 1/(n+1) to simplify our calculation.

1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(98x99) + 1/(99x100) = [1/1 - 1/2] + [1/2 - 1/3] + [1/3 - 1/4] + ... + [1/98 - 1/99] + [1/99 - 1/100] = 1/1 - 1/100 = 99/100 = 0.99

Notice that -1/2 and +1/2 cancel up each others, so do -1/3 and -1/3, and the other pairs. These are called additive inverses. The sum of any additive inverse with its pair will yield zero as the answer.
10. A = the number of sides of a pentagon; B = the third odd positive integer; C = the fifth Fibonacci number (the sequence starts with 1, 1, 2...); D = the number of diagonals of a pentagon; E = the third prime number. What is the sum of A, B, C, D and E? In other words, compute A + B + C + D + E.

Answer: 25

A pentagon has 5 sides and also 5 diagonals. So, A = D = 5. The first few odd numbers are 1, 3, 5, 7... So, B = 5. The first few Fibonacci numbers are 1, 1, 2, 3, 5... Hence, C = 5. The first few prime numbers are 2, 3, 5, 7, 11...Therefore, E = 5.

A + B + C + D + E = 5 + 5 + 5 + 5 + 5 = 5 x 5 = 25
Source: Author Matthew_07

This quiz was reviewed by FunTrivia editor crisw before going online.
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