Advanced Math/25


1. An astronomer claims that there are one hundred thousand million galaxies in the universe, each containing one hundred thousand million stars.  If you write the total number of stars this would mean as a power of ten, A = the power of ten.

2. If x+y+z=1, x+y-z=2 and x-y-z=3, B= 10*x*y*z (where * means multiply)

3. Let N be the smallest positive integer whose digits add up to 2013.  C= the first digit of N+1.

4. The number 3 can be expressed as the sum of one or more positive integers in four different ways: 3, 1+2, 2+1, 1+1+1.  D= the number of different ways in which the number 5 can be expressed.

5. The equation x^2+y*x+z=0, where y and z are different, has solutions x=y and x=z.  E=5y+z.

I've made an attempt at these. I have A=11, B=5 and D=8, which I'm reasonably confident of. However, the other two have me beat.

The questions are a preliminary step to a geocaching puzzle. Once I have values for A to E I use them in the next stage (which I think I will be able to manage for myself). 40 years ago I might have managed without help but I've forgotten so much!

Question 3 requires you to consider how the digits can be distributed in such a way that they produce the smallest number possible. Let's say you wanted to add up to 18. Would you rather use 9s or 1s?

99 → 9+9 = 18
111111111111111111 → 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = 18

Obviously, 111111111111111111 is much bigger than 99.

In order to minimize the number, we will use as many 9s as possible. This will allow us to use the fewest total digits, which makes the number as small as it could be.

2013 / 9 gives you 223 with remainder 6, so the number you want is 6 followed by 223 9s.


Of course, it's not asking for the six -- it wants you to add 1 to this number! That regroups all the way to give you 710^23, so C=7.

Question 5 needs you to solve for y and z. You can eliminate x by plugging it in:

x = y → y^2 + y^2 + z = 0

x = z → z^2 + y z + z = 0

If both of those things are true, can you solve for y and z? The three possible solutions are (-1/2,-1/2), (0,0), and (1,-2), which give you the values -3, 0, and 3 respectively for 5y+z. You take the one where y and z are different, so (1,-2) gives E=3.

Also, double check your answers for A and D. Your A should be 22 I believe (it's 100*1000*1000000)^2 not just 100*1000*1000000). Your answer for D should be 16, as there are the following:


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Clyde Oliver


I can answer all questions up to, and including, graduate level mathematics. I am more likely to prefer questions beyond the level of calculus. I can answer any questions, from basic elementary number theory like how to prove the first three digits of powers of 2 repeat (they do, with period 100, starting at 8), all the way to advanced mathematics like proving Egorov's theorem or finding phase transitions in random networks.


I am a PhD educated mathematician working in research at a major university.


Various research journals of mathematics. Various talks & presentations (some short, some long), about either interesting classical material or about research work.

BA mathematics & physics, PhD mathematics from a top 20 US school.

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Various honors related to grades, various fellowships & scholarships, awards for contributions to mathematics and education at my schools, etc.

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In the past, and as my career progresses, I have worked and continue to work as an educator and mentor to students of varying age levels, skill levels, and educational levels.

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