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Question
1) A warehouse contains 200 shoes of size 8, 200 shoes of size 9 and 200 shoes of size 10. Of these 600 shoes, there are 300 left shoes and 300 right shoes. What is the minimum number of usable shoes?
(a) 50 (b) 100 (c) 200 (d) none of these

2) If 'K' is any natural number, such that 100<=K<=200, how many values of K exist such that K! has 'z' zeroes at its end and (K+2)! has 'z+2' zeroes at its end?
(a) 2 (b) 4 (c) 6 (d) none of these

3) There exists a 5 digit number N with distinct and non-zero digits such that it equals the sum of all distinct three digit numbers whose digits are all different and are all digits of 'N'. Then the sum of the digits of 'N' is a necessarily?
(a) Perfect Square (b) Cube (c) Even (d) None of these

4) Starting with 1, positive integers are written one after the other. What is the 40,0000th digit that will be written?
(a) 3 (b) 6 (c) 8 (d) none of these

5) If in the number system of a particular country, 25 means 5 tens and 2 units, 467 means 7 hundreds, 6 tens and 4 units. Then find the value of 173 x 425? (x here means multiplication)
(a) 4,04,491 (b) 7,35,255 (c) 6,22,744 (d) 5,25,376

If 200 left were size 10 and 200 right were size 8, that would leave 200 that were size 9,
and I believe that is the minimum number of usable shoes, so (c).

2) For this to occur, there would have to be two numbers between 100 and 200 that gave 2 more zeroes to the end of the number.  Now 100 added two more zeroes, but if that were K+2, then K would be 98, and that's  not in the realm.  That means there is only one such number: K=198.
This puts K+2 at 200, which adds 2 more zeroes.  If K=199, K+2 would be outside of the domain.
From that, there is only one such number, so that answer is (d).  It is (a) at first glance, for 100 adds two zeroes, but 100 can't be K+2.  If 98, 99, and 199 were allowed in as well, then there would be 4 numbers total, and that is (b).

3) All that can said about this sum is that it is 6 times the sum of the original numbers.
This is not a square, it is not a cube, but it is even.  This means the answer is (c).

4) The first 9 numbers (1-9) have 1 digit each, for a total of 9 digits written.  The next 90 numbers (10-99) have 2 digits each, for 180 digits, and a total of 189 digits.  The next 900 numbers (100-999) have 3 digits each, for 2,700 digits, and a total of 2,889 digits.  The next 9000 numbers (1000-9999) have 4 digits each, for 36,00 digits, for a total of 38,889.

This means the digit in question is 1,111 digits into the next set.  Since each number has 5 digits, that mean that there have been 1,110 digits written down.  This means that it is the 1st digit in the number after that.  Since 1000 numbers get us into the two thousands, the first digits in this number must be a 2 since 1,110 is only 110 greater than 1,000.
Again, I get none of these.

5) This means that 25 is read to really be 52 and 467 is really read to be 764.  This means the numbers 173 and 425 area really 371 and 524.  Multiplying these two gives 194,404.
This means the answer is (a), since that's the result of reversing the digits in 194,404.

Volunteer

#### Scott A Wilson

##### Expertise

I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can also say that I broke 5 minutes for a mile, which is over 12 mph, but is that relevant?

##### Experience

Experience in the area; I have tutored people in the above areas of mathematics for over two years in AllExperts.com. I have tutored people here and there in mathematics since before I received a BS degree back in 1984. In just two more years, I received an MS degree as well, but more on that later. I tutored at OSU in the math center for all six years I was there. Most students offering assistance were juniors, seniors, or graduate students. I was allowed to tutor as a freshman. I tutored at Mathnasium for well over a year. I worked at The Boeing Company for over 5 years. I received an MS degreee in Mathematics from Oregon State Univeristy. The classes I took were over 100 hours of upper division credits in mathematical courses such as calculus, statistics, probabilty, linear algrebra, powers, linear regression, matrices, and more. I graduated with honors in both my BS and MS degrees. Past/Present Clients: College Students at Oregon State University, various math people since college, over 7,500 people on the PC from the US and rest the world.

Publications
My master's paper was published in the OSU journal. The subject of it was Numerical Analysis used in shock waves and rarefaction fans. It dealt with discontinuities that arose over time. They were solved using the Leap Frog method. That method was used and improvements of it were shown. The improvements were by Enquist-Osher, Godunov, and Lax-Wendroff.

Education/Credentials
Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

Awards and Honors
I earned high honors in both my BS degree and MS degree from Oregon State. I was in near the top in most of my classes. In several classes in mathematics, I was first. In a class of over 100 students, I was always one of the first ones to complete the test. I graduated with well over 50 credits in upper division mathematics.

Past/Present Clients
My clients have been students at OSU, people who live nearby, friends with math questions, and several people every day on the PC. I would guess that you are probably going to be one more.