A is the 1^st number in the alphabet

J is the 10^th number in the alphabet

\(\Rightarrow\) \(\dfrac{2}{5}\) of 10 (from A to J) is the 4^th number in the alphabet, which is D

In the first column, the sum of the digits of every number is 12

In the second column, the sum of the digits of every number is 10

In the third column, the sum of the digits of every number is 6

Maybe, the missing number is 60, because I can't find any number which has the sum of its digits is 6 and different from other numbers!

If 3x - 1 = 0 \(\Leftrightarrow\) x = \(\dfrac{1}{3}\) \(\Rightarrow\) The equality is always correct

If 3x - 1 \(\ne\) 0 \(\Rightarrow\) x² + 2 = 7x - 10

\(\Rightarrow\) x² - 7x + 2 = -10

\(\Leftrightarrow\) x² - 2 \(\times\) 3,5 \(\times\) x + (3,5)² - (3,5)² + 2 = -10

\(\Leftrightarrow\) (x - 3,5)² - 10,25 = -10

\(\Leftrightarrow\) (x - 3,5)² = 0,25 = (0,5)²

\(\Rightarrow\) \(\left[{}\begin{matrix}x-3,5=0,5\\x-3,5=-0,5\end{matrix}\right.\)\(\Rightarrow\) \(\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\) (3x - 1 \(\ne\) 0)

So x \(\in\) {\(\dfrac{1}{3}\);3;4}

in units, the number 4 appears 10 times (404, 414, 424, ..., 494)

in tens, the number 4 appears 10 times (440 - 449)

in hundreds, the number 4 appears in every number from 400 to 499, which means there are 100 numbers 4 in hundreds.

In total, we have to write: 10 + 10 + 100 = 120 times number 4

What a trick! Look carefully, you might see, without space: 135791113151

Or: 1, 3, 5, 7, 9, 11, 13, 15, ...

So the number, which replace the question mark, is 79

\(\)

One can of dog food can feed 8 pups or 6 dogs, so we can assume 8p = 6 d

With 8 cans, we can feed 8 × (8p) = 56p

We have: 56p = 40p + 16p = 40p + 12d

So 8 cans can feed 40 puppies and 12 dogs

(Without your clue, I think I will be lost in this puzzle!)

A million has 6 zeros: 1,000,000

A thousand has 3 zeros: 1,000

In total, a thousand thousand million has: 6 + 3 × 2 = 12 (zeros)

HOW BIG: 1,000,000,000,000

We have: (a₁ + a₂) + (a₂ + a₃) + ... + (a₅₁ + a₁) = 1 + 2 + ... + 51

\(\Leftrightarrow\) 2(a₁ + a₂ + ... + a₅₁) = (51 + 1) × 51 ÷ 2 = 1326

\(\Leftrightarrow\) a₁ + a₂ + a₃ + ... a₅₁ = 1326 ÷ 2 = 663

Mean: (2 + 0 + 1 + 3 + 0 + 3 + 3 + 1 + 2) ÷ 9 = 5/3

From the smallest to largest: 0; 0; 1; 1; 2; 2; 3; 3; 3

\(\Rightarrow\) The median is 2 (in the middle)

Mode: 3 (repeated 3 times)

We have: 5/3 < 2 < 3 \(\Rightarrow\) mean < median < mode

ANSWER: C

x - x² - 1 = -(x + x² + 1)

= -[x² + 2x\(\dfrac{1}{2}\) + (\(\dfrac{1}{2}\))²+ \(\dfrac{3}{4}\)]

= -[(x + \(\dfrac{1}{2}\))² + \(\dfrac{3}{4}\)]

(x + \(\dfrac{1}{2}\))² \(\ge\) 0 \(\Rightarrow\)(x + \(\dfrac{1}{2}\))² + \(\dfrac{3}{4}\) > 0

\(\Rightarrow\) -[(x + \(\dfrac{1}{2}\))² + \(\dfrac{3}{4}\)] < 0

\(\Leftrightarrow\) x - x² - 1 < 0

what about 9?

I think you should say: f(x) is divided by ..., because f(x) is divisible when it doesn't have any remainder.

11 inches tall after burning for 4 hours

8 inches tall after burning for 6 hours

\(\Rightarrow\) Every 6 - 4 = 2 hours the candle was 11 - 8 = 3 inches shorter

\(\Rightarrow\) After 4 hour the candle was 6 inches shorter

\(\Rightarrow\) The candle was: 11 + 6 = 17 (inches) tall

Let b be the number of bunnies (b \(\in\) N* and b < 27)

\(\Rightarrow\) There are 27 - b chicks

\(\Rightarrow\) 4b + 2(27 - b) = 78 (legs)

\(\Leftrightarrow\) 2b = 24

\(\Leftrightarrow\) b = 12

Hence, there are 12 bunnies

P/s: This is the second solution for this math problem. The first one has been worked out by Trịnh

Let x (mi/h) be the average speed driving from starting location to the delivery location.

\(\Rightarrow\) The average speed driving from delivery location(DL) to the starting location(SL) is x - 20 (mi/h)

It took her 180/x (hours) to travel from SL to DL and 180/(x - 20) (hours) to travel from DL to SL, which means: 180/x + 180/(x - 20) = 7,5

\(\Rightarrow\) x = 60 

So her average driving from SL to DL is 60 mi/h

We have: 0 \(\times\) (-1) \(\times\) k = 0 - an element in the set

\(\Rightarrow\) With k \(\in\) R, the set {0, -1, k} is always closed under multiplication

The numbers which are written as the sum of two prime numbers are:

4 = 2 + 2

5 = 2 + 3

6 = 3 + 3

7 = 2 + 5

8 = 3 + 5

9 = 7 + 2

10 = 5 + 5

12 = 7 + 5

13 = 11 + 2

14 = 7 + 7

15 = 13 + 2

16 = 13 + 3

18 = 11 + 7

19 = 17 + 2

So the numbers left are: 1, 2, 3, 11, 17

The sum of all positive integers less than 20 that cannot be written as the sum of two prime numbers is:

1 + 2 + 3 + 11 + 17 = 34

We have: 5 = 1 + 0 + 4 = 1 + 1 + 3 = 1 + 2 + 2 = 2 + 0 + 3 = 5 + 0 + 0

With 1, 0, 4, we can have 4 three-digit numbers including these 3 numbers

With 1, 1, 3, we can have 6 three-digit numbers including these 3 numbers

With 1, 2, 2, we can have 6 three-digit numbers including these 3 numbers

With 2, 0, 3, we can have 4 three-digit numbers including these 3 numbers

With 5, 0, 0, we can have 1 three-digit number including these 3 numbers

In total, we have 4 + 6 + 6 + 4 + 1 = 21 (numbers)

a - b = 8 => a = 8 + b

=> ab = (b + 8)b = 48

By replacing b by 1, 2, 3, 4, ... respectively, we have b = 4

=> a = 12 => a + b = 12 + 4 = 16

x - y = 1 \(\Rightarrow\) y = x - 1 (y < -1)

\(\Rightarrow\) xy = y(y - 1) = y² - y

y < 0 \(\Rightarrow\) y = -z (z > 0)

\(\Rightarrow\) xy = (-z)² - (-z) = z² + z

xy has the minimum value \(\Leftrightarrow\) z² + z has the minimum value ​\(\Leftrightarrow\) z has the minimum value \(\Leftrightarrow\) y has the maximum value \(\Leftrightarrow\) y = -2 \(\Leftrightarrow\) x = -1