The document contains solutions to multiple math problems. The first problem finds the smallest natural number that when multiplied by 123 yields a product ending in 2004. The second problem involves determining the value of x in a geometry problem where two angles are equal. The third problem involves simplifying an algebraic expression by factorizing differences of squares.

1. Find the smallest natural number which when multiplied by 123 will yield a
product that ends in 2004
Solution:
Assume 𝒙 is the smallest natural number
𝟏𝟐𝟑𝒙 = 𝑨 × 𝟏𝟎 𝟒 + 𝟐𝟎𝟎𝟒 ---- ①
When 𝑨 = 𝟎, 𝟐𝟎𝟎𝟒 is not divisible by 123
Answer∶ 𝒙 = _____
When 𝑨 = 𝟏, 𝟏𝟐𝟎𝟎𝟒 is not divisible by 123
When 𝑨 = 𝟐, 𝟐𝟐𝟎𝟎𝟒 is not divisible by 123
When 𝑨 = 𝟑, 𝟑𝟐𝟎𝟎𝟒 is not divisible by 123
When 𝑨 = 𝟒, 𝟒𝟐𝟎𝟎𝟒 is not divisible by 123
When 𝑨 = 𝟓, 𝟓𝟐𝟎𝟎𝟒 is not divisible by 123
When 𝑨 = 𝟔, 𝟔𝟐𝟎𝟎𝟒 is not divisible by 123
When 𝑨 = 𝟕, 𝟕𝟐𝟎𝟎𝟒 is not divisible by 123
When 𝑨 = 𝟖, 𝟖𝟐𝟎𝟎𝟒 is not divisible by 123
When 𝑨 = 𝟗, 9𝟐𝟎𝟎𝟒 ÷ 𝟏𝟐𝟑 = 𝟕𝟒𝟖

2. In the following figure, if CA = CE, what is the value of 𝑥 ?
Solution:
𝒙°
32° 36°
6𝟖°
6𝟖°
Answer∶ 𝒙° = 𝟏𝟖𝟎° − 𝟔𝟖° − 𝟔𝟖° = _____°
∠𝑪𝑬𝑨 = ∠𝑬𝑪𝑫 + ∠𝑬𝑫𝑪 = _____°
𝑪𝑨 = 𝑪𝑬 ⇒ ∠𝑪𝑨𝑬 = ∠𝑪𝑬𝑨 = _____°

3. Solution:
The above question can be re-written as:
20052 − 20042 + 20032 − 20022 + … − 42 + 32 − 22 + 12
Compute: 12 − 22 + 32 − 42 + … − 20022 + 20032 − 20042 + 20052
We know that
𝑎2
− 𝑏2
= 𝑎 + 𝑏 (𝑎 − 𝑏)
Answer: 2011015
= 2005 + 2004 2005 − 2004 + 2003 + 2002 2003 − 2002 + …
+ 3 + 2 3 − 2 + 12
= 2005 + 2004 + 2003 + 2002 + … + 3 + 2 + 1
Sum of sequence with common difference:
𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚 + 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 × 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠
2
=
2006 × 2005
2

4. Solution:
After 15 minutes, which is 5:45am
M has travelled = 100 ×
15
60
= ___ 𝑘𝑚
City P is 625 kilometers from City Q. M departed from City P at 5.30am
travelling at 100 kilometers per hour, and arrived at City Q. Fifteen minutes
after M left, N departed from City Q and arrived at City P travelling at 80
kilometers per hour. At what time did M and N meet together?
The distance between M and N = 625 − ___ 𝑘𝑚 ---- ①
The time needed to meet =
①
100+80
=
10
3
hours
∴ The time they meet is 5:45am +
10
3
hours
Answer: ________________________________