- 5 kids A, B, C, D, E are sitting around a table with candies amounts of 10, 30, 20, 20, 40 - In each round, each kid gives half their candies to the kid on their right - If a kid ends with an odd amount, they take 1 more from the table - The question is if after several rounds all kids can have the same amount, and if so how much

1. Solution:
Let the number be 𝑎𝑏𝑐𝑑𝑒 and 𝑎, 𝑏, 𝑐, 𝑑, 𝑒 are all different and > 0
𝑎𝑏𝑐𝑑𝑒 = multiple of 4  𝑑𝑒 is multiple of 4
𝑎𝑏𝑐𝑑 = multiple of 5  𝑑 is 5, and 𝑑𝑒 is either 52 or 56
𝑎𝑏𝑐 = multiple of 6
𝑎𝑏 = multiple of 7  𝑎𝑏 can be 14, 21, 28, 42, 49, 63, 84, 91, 98
(35,56 cannot because d is 5, 70 cannot because all digits > 0, 77 cannot
because all digits are different)
𝑎𝑏𝑐 = multiple of 6  𝑎𝑏𝑐 can be 216, 426, 492, 498, 846, 912, 918, 984
(144, 282, 288, 636 cannot because all digits are different, 210, 630 cannot
because all digits > 0, )
There is a five-digit number with all the digits different and all are greater
than zero. It is a multiple of 4. If the ones digit is cancelled, then the four-
digit number left will become a multiple of 5. If the ones digit of the
newly-formed four-digit number is cancelled again, then the three-digit
number left will become a multiple of 6. If the ones digit of this three-digit
number is cancelled again, the number formed by this two-digit number
will become multiple of 7. How many such kind of five-digit numbers are
there?
So we have:
𝑎𝑏𝑐 can be 216, 426, 492, 498, 846, 912, 918, 984
𝑑𝑒 is either 52 or 56
Possible number: 49256, 49852, 49856, 84652, 91256, 91852, 91856, 98452,
98456
Answer: 9

2. Four friends are lined-up in a row from the oldest to the youngest such
that the difference of every two of the neighbouring ages in the row is the
same and the product of their ages is 1680, what is the sum of their ages?
Solution:
Let their age be a, b, c, d, and a > b > c > d.
Given a – b = b – c = c – d = k
And a x b x c x d = 1680
First, find the prime factorization for 1680:
1680 = 2 x 2 x 2 x 2 x 3 x 5 x 7
= 5 x 6 x 7 x 8 or = 2 x 6 x 10 x 14
From above, we can tell
a = 8, b = 7, c = 6, d = 5
or a = 14, b = 10, c = 6, d = 2
a + b + c + d = 8 + 7 + 6 + 5 = ___
or a + b + c + d = 14 + 10 + 6 + 2 = ___
Answer:26 𝑜𝑟 32

3. The sum of all the remainders of 3 consecutive two-digit numbers when
divided by 5 is 7. The sum of all the remainders of those numbers when divided
by 7 is 9, and the sum of all the remainders of those numbers when divided by
9 is 15. What is the sum of the remainders of those numbers when divided by
11?
Solution:
Let the 3 consecutive two-digit numbers as x, y, z
The possible remainders for 5: 1, 2, 3, 4, 0
The possible remainders for 7: 1, 2, 3, 4, 5, 6, 0
The possible remainders for 9: 1, 2, 3, 4, 5, 6, 7, 8, 0
Since x, y, and z are consecutive numbers, their remainder are also ‘consecutive’
Given: x y z
sum of remainder divided by 5 is 7, the three remainders are: 3, 4, 0
sum of remainder divided by 7 is 9, the three remainders are: 2, 3, 4
sum of remainder divided by 9 is 15, the three remainders are: 4, 5, 6
Look at the ‘z’ column: since the remainder is 0 when divided by 5, z is multiple
of 5.
And when divided by 7, it has a remainder of 4. we can figure that possible
values for z are 25 or 60 or 95.
Also it must fulfill the condition that when divided by 9, the remainder is 6. Only
60 fulfill this condition.
Therefore, 𝑥 = 58, 𝑦 = 59, 𝑧 = 60
Answer: the sum of remainders divided by 11 is 12

4. Solution:
Five kids A, B, C, D and E are sitting around a circular table (refer to the
given figure) with some candies. Each of them gets an even number of
candies. The quantities are 10, 30, 20, 20 and 40 respectively. In the first
round, each of them gives one half of their candies to the kid to their right.
At this time, the amounts of their candies become 25, 20, 25, 20 and 30
respectively. If the kid’s number of candies is odd, then he/she should
pick one from the table. Is it possible that the kids have the same number
of candies after several rounds? How many pieces does everyone have?