Your problem is not true, because 3210 = 2 x 3 x 5 x 107, and noone in this problem can be 107 years old!

We have: 6:10 pm = 18:10.

Since a day has 24 hours, we can rewrite 8:20 am the next morning as 32:20 of the same day as 18:10.

So there are 32:20 \(-\) 18:10 = 14 hours 10 minutes between 6:10 pm and 8:20 am the next morning.

Answer. 14 hours 10 minutes

There are 24 hours in a day.

So, a quarter of a day = 24 \(\times\dfrac{1}{4}\) = 6 hours.

Thus, a third of 6 hours = 6 \(\times\dfrac{1}{3}\) = 2 hours.

Thereupon (also means Thus), a half of 2 hours = 2 \(\times\dfrac{1}{2}\) = 1 hour.

Therefore, a half of a third of a quarter of a day is 1 hour. So we choose C.

Answer. C

Let 140a be the number of refrigerators Alvin sold in 2009, where a is a constant value. Thus,

- Ben sold 140a \(\div\) 7 = 20a (refrigerators);

- Carl sold 140a \(\div\) 5 = 28a (refrigerators);

- Dan sold 140a \(\div\) 4 = 35a (refrigerators).

According to the problem, we have,

140a + 20a + 28a + 35a = 669

223a = 669

a = 3.

Therefore, Alvin had sold at most 3 \(\times\) 140 = 420 refrigerators.

Answer. 420 refrigerators

Let [a,b] be the LCM of a and b, and (a,b) be the GCD of a and b.

Since (a,b) = 8, [a,b] = 96, then a \(\times\) b = [a,b] \(\times\) (a,b) = 96 \(\times\) 8 = 768.

As (a,b) = 8, let a = 8m, b = 8n, in which (m,n) = 1.

=> a \(\times\) b = 8m \(\times\) 8n = 768. Thus, m \(\times\) n = 768 \(\div\) (8 \(\times\) 8) = 12.

(m,n) = 1, so all the pairs (m;n) which have GCD = 1 and m \(\times\) n = 12 are:

(1;12),(3;4),(4;3),(12;1).

=> (a;b) = (8;96),(24;32),(32;24),(96;8).

Therefore, there are 2 possible values of a + b: 8+ 96 = 104, 24 + 32 = 56.

Answer. 104 and 56

The number has ending digits 0, 2, 4, 6, 8. Good luck!

Nguyễn Việt Hoàng has copied my answer!

||4x-2| - 2| = 4

Scenerio (S.c.) 1:

|4x - 2| - 2 = 4

|4x - 2| = 6

S.c. 1.1:

4x - 2 = 6

4x = 8

x = 2.

S.c. 1.2:

4x - 2 = -6

4x = -4

x = -1.

S.c. 2:

|4x - 2| - 2 = -4

|4x - 2| = -2

|4x - 2| is always larger or equal to 0, so there is no value of x in this scenerio.

Thus, the values of x are 2; -1.

From 7m + 5n = 31, we can see that 7m or 5n is even.

- If 7m is even, m is even and prime => m = 2.

=> 5n = 31 - 7 x 2 = 17. n \(⋮̸\) 5, so no numbers sastify the problem.

- If 5n is even, n is even and prime => n = 2.

=> 7m = 31 - 5 x 2 = 21 => m = 3 is a prime number.

The value of:

mⁿ + n^m = 3² + 2³ = 9 + 8 = 17.

1339 = 103 x 13 = 1 x 1339.

No one can be 1339 years old, so the grandfather will be 103 years old and his grandchild will be 13 years old next year.

Thus, the grandfather is 102 years old, and his grandchild is 12 years old now.

100 + 500 : 5 - 40

= 100 + 100 - 40

= 200 - 40 

= 160.

5n8 is divisible by 9 => (5 + n + 8) \(⋮\) 9 means (13 + n) \(⋮\) 9.

=> n = 5.

=> 3m9 = 558 - 189 = 369 => m = 6.

So, m + n = 5 + 6 = 11.

abcabc = abc \(\times\) 1001

            = abc ​ ​\(\times\) 91 \(\times\) 11

Thus, abcabc is divisible by 11.