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 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 Answer :𝒙 = _____

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

Compute: 1 2 − 2 2 + 3 2 − 4 2 + …− 2002 2 + 2003 2 − 2004 2 + 2005 2 Solution: The above question can be re-written as: 2005 2 − 2004 2 + 2003 2 − 2002 2 + …− 4 2 + 3 2 − 2 2 + 1 2 We know that 𝑎 2 − 𝑏 2 = 𝑎+𝑏 (𝑎 −𝑏) = 2005+2004 2005−2004 + 2003+2002 2003−2002 + … + 3+2 2−2 + 1 2 = 2005+2004 + 2003+2002 + … + 3+2 +1 = 2006 ×2005 2 Sum of sequence with common difference: 𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚+𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 ×𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠 2 Answer: 2011015

City P is 625 kilometers from City Q. M departed from City P at 5 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? Solution: After 15 minutes, which is 5:45am M has travelled = 100 × 15 60 =___ 𝑘𝑚 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: ________________________________