Weekly Dose 20 - Maths Olympiad Practice

1. Solution:
From 𝐶 + 𝐼 = 𝐶, and since 𝐼 cannot be 0, 𝐼 can either be 7, 8 or 9.
Then from 𝐶 + 𝐶 + 𝐶 + 𝐶 = 𝐼, 𝐼 can only be 8. No multiple of 4 ends with 7 or
9.
Alphametric is a puzzle which replaces numerical digits in arithmetic
problems with letters. Each letter represents a digit and no digits are
represented by two different letters. What does the three-digit number
𝐼𝑀𝐶 equal to ?
Let’s try 𝐶 = 2:
Because 2 + 8 = 𝑀2, then 2 + 8 + 𝐸 + 𝐸 𝑚𝑢𝑠𝑡 𝑏𝑒 22 ,
thus 𝐸 = 6, M = 1, S = 4
Answer: 812
As such, 𝐶 can either be 2 or 7.

2. Neal uses the month of his birthday and multiply it by 31; then the day by
12. He then adds the two products and get a sum of 400. On what month
and day is Neal’s birthday?
Solution:
Let 𝑚 be the birthday month and 𝑑 be the birthday day,
31𝑚 + 12𝑑 = 400, 1 ≤ 𝑚 ≤ 12, 1 ≤ 𝑑 ≤ 31
12𝑑 always returns even number, 400 is also an even number, therefore 31𝑚
must be an even number as well.
To get an even number out of 31𝑚, 𝑚 must be even.
Also, since 12𝑑 and 400 are multiple of 4, 31𝑚 must be multiple of 4 as well,
Which means 𝑚 can either be 4, or 8 or 12
Since 𝑑 must be natural number ..
Answer: April 23
𝑚 31𝑚 12𝑑 = 400 – 31𝑚 𝑑
4 124 276 23
8 248 152 12.67
12 372 28 2.33

3. The sum of two different numbers is 43.67. If the decimal point of the smaller
number moves one place towards the right side, then it will equal to the bigger
one. What is the bigger number.
Solution:
Assume the two numbers are: 𝑠 (for smaller one), and 𝑏
Given 𝑠 + 𝑏 = 43.67
If 𝑠 = 𝑎𝑏. 𝑐𝑑, then 𝑏 = 𝑎𝑏𝑐. 𝑑 , but 𝑎𝑏. 𝑐𝑑 + 𝑎𝑏𝑐. 𝑑 ≠ 43.67, so 𝑠 ≠ 𝑎𝑏. 𝑐𝑑
If 𝑠 = 𝑎. 𝑏𝑐, then 𝑏 = 𝑎𝑏. 𝑐 , then it is possible that 𝑎. 𝑏𝑐 + 𝑎𝑏. 𝑐 = 43.67
From above, we can derive from right:
𝑐 = 7,
𝑏 = 9,
𝑎 = 3
Answer: 39.7

4. Solution:
2012 pupils are arranged in one row and are supposed to call numbers
according to a certain pattern. If one student calls a one-digit number,
then the student next to him must call a number twice of it. When a
student call a two-digit number, then the next pupil after him must call the
sum of 8 and the ones digit of that two-digit number. If the first pupil calls
1, then what number must the last pupil in the row call?
Pattern starts at 4th call, 2012 – 3 = 2009
2009 ÷ 5 = 401 … 4
Remaining is 4, therefore the last pupil call is: ___
Answer: 12