# Tear the question paper, then throw it away

After the APMOPS, here are some questions that I translated:

1. A magician made a six-digit number $A$, and the digit sum of $A$ be $B$. The magician called out a spectactor to evaluate $A$-$B$. The spectactor said out 5 numbers, namely $0,2,4,6$ and $8$.The magician successfully revealed the last number. What number is it? Explain your reason.

*Bonus question: Find the minimum and maximum value for $A$.

1. The eight vertexes of an octagon are attached with circles, each required to fill up the numbers from 1 to 8. Can the sum of four consecutive attaching circles be:

a)larger than 16?

b) larger than 17?

If possible, find a way of doing so; if not, explain your reason. 1. Wong cycled from station A to station B. Buses form station A and B each give out a bus at the same interval time (e.g. when station A gives out a bus every 30 minutes, station B does the same), but at different times. Every 6 minutes Wong meets up with a bus coming from the opposite, and every 9 minutes he is overtaken by a bus traveling at the same direction with him. It is known that all the buses from stations A & B take 50 minutes to travel to the other side ( it means that buses from station A travel 50 minutes to station B, and vice versa ). How long does it take for Wong to travel from station A to B?

4.In the figure below, 3 different heights of the triangle move from the bases to the vertexes of the triangle. P is a point in the triangle such that another 3 lines move from point P to the triangle, causing each of the lines to be parallel with AD, BE and CF respectively. If AD=2010 cm, BE= 2013 cm, CF = 2016 cm and PR = 1005 cm, PS= 671 cm, find the length of PT. (Oh, and I forgot, each line extending to the base is straight, or 90 degrees) Feel free to discuss! Enjoy! Note by Bryan Lee Shi Yang
4 years, 6 months ago

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Q3 is 250 minutes.

- 6 months ago

Q4 is 336.

- 6 months ago

Q1: The last number is $7$. $A$ can be expressed as $(10^5\times a+10^4\times b+10^3\times c+10^2\times d+10\times e+f)$ where $a,b,c,d,e$ and $f$ are the digits of $A$.

$B$ can be expressed as $(a+b+c+d+e+f)$.

Now, $A-B=99999a+9999b+999c+99d+9e$.

$A-B=9(11111a+1111b+111c+11d+e)$ implying that $A-B$ is a multiple of $9$.

Since, $A-B$ is a multiple of $9$, the digit sum of $A-B$ must also be divisible by $9$ (Divisibility Test of $9$).

Now, we are provided with $5$ digits of $A-B$ and we have to find the sixth one. The sum of these $5$ digits comes out to be $20$ and so for $A-B$ to be divisible by $9$, the sixth digit should be $7$ so that it sums out to be a multiple of $9$.

NOTE: There is a trick here. If the spectator would have called out $2,4,6,7$ and $8$ and then we were to find out the sixth digit, we could have $2$ different answers namely $0$ and $9$. Can you figure it out why?

- 4 years, 6 months ago

Because 2+4+6+7+8 IS a multiple of 9

- 4 years, 6 months ago

Yes, it has to be a multiple of $9$. This is true for all numbers and can be generalized easily for $n$-digit numbers.

- 4 years, 6 months ago