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 AA, and the digit sum of AA be BB. The magician called out a spectactor to evaluate AA-BB. The spectactor said out 5 numbers, namely 0,2,4,60,2,4,6 and 88.The magician successfully revealed the last number. What number is it? Explain your reason.

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

  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
6 years, 3 months ago

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Q3 is 250 minutes.

Saya Suka - 2 years, 3 months ago

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Q4 is 336.

Saya Suka - 2 years, 3 months ago

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Q1: The last number is 77. AA can be expressed as (105×a+104×b+103×c+102×d+10×e+f)(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,ea,b,c,d,e and ff are the digits of AA.

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

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

AB=9(11111a+1111b+111c+11d+e)A-B=9(11111a+1111b+111c+11d+e) implying that ABA-B is a multiple of 99.

Since, ABA-B is a multiple of 99, the digit sum of ABA-B must also be divisible by 99 (Divisibility Test of 99).

Now, we are provided with 55 digits of ABA-B and we have to find the sixth one. The sum of these 55 digits comes out to be 2020 and so for ABA-B to be divisible by 99, the sixth digit should be 77 so that it sums out to be a multiple of 99.

NOTE: There is a trick here. If the spectator would have called out 2,4,6,72,4,6,7 and 88 and then we were to find out the sixth digit, we could have 22 different answers namely 00 and 99. Can you figure it out why?

Yash Singhal - 6 years, 3 months ago

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

Bryan Lee Shi Yang - 6 years, 3 months ago

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Yes, it has to be a multiple of 99. This is true for all numbers and can be generalized easily for nn-digit numbers.

Yash Singhal - 6 years, 3 months ago

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