Mixed Practice Level 3 Exams


Mathcount Mixed Practice Level 3 Exams


Level 3 mixed practice level is much advance practice set that helps students think outside the box with a various different level and different pattern question sets.Proficient training is provided by the best in industry trainers who provide in-depth practice sessions and training material. It helps develop skills to solve problems within limited time framed and short cuts that help resolve set of questions easily.



Here are a few examples of course problems with their solutions.

Question There is at least one two-digit number such that when it is added to the two-digit number having the same digits in reverse order, the sum is a perfect square. Find the sum of all such two-digit numbers.
Answer Let 10a + b be the number. Then 10b + a is the same number with its digits reversed. 10a + b + 10b + a = 11a + 11b = 11(a + b) If this number is a perfect square, a + b = 11. Since a and b are single digits £ 9, a can take the range from 2 to 9, bit b taking the range from 9 to 2. 29 + 38 + 47 + 56 + 65 + 74 + 83 + 92 = 10(2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 10(44) + 44 = 11(44) = 484 Answer

Problem Call a multi-digit positive integer divisorly if, for each pair of adjacent digits Aand B, either A = Bn or B = An for some integer n. For example, 12639 is divisorly because 1 evenly divides 2, 2 evenly divides 6, 3 evenly divides 6, and 3 evenly divides 9. What is the greatest divisorly integer whose digits are all different?
Solution We would like to be able to use all 10 digits in our number, and the leftmost digit should be a 9, if possible, to achieve the largest qualifying 10-digit number. Let’s start with 9. The next digit needs to divide 9 or be divisible by 9—only 3, 1 and 0 work, so try the largest, 3. Then the next digit needs to divide 3 or be divisible by (and not already be used)—only 6, 1 and 0 work, so try the largest, 6. Then the next digit must be 2, 1 or 0, so try the largest, 2. The next digit needs to be 8, 4, 1 or 0, so try the largest, 8. The next digit needs to be 4, 1 or, 0, so try the largest, 4. The next digit needs to be 1 or 0, so try the largest, 1. Since 1 divides anything, it will work we whatever we have left: 7, 5 or 0, so try the largest, 7. Only 0 and 5 remain, and, of these two, 7 divides only 0, so that must come next. For the last digit, 5 will work, because 5 divides 0. We have successfully completed the largest possible qualifying number: 9,362,841,705.


The sum of the digits of a positive, two-digit integer is 9. The positive difference between the integer created when the two digits are reversed and the original integer is 27. What is the product of these two digits? Answer :18 Solution : The sum of the digits of a two-digit positive integer is 9. The difference between this integer and a two-digit integer containing the digits reversed is 27. We must find the product of the two digits. Let x and y be the two digits. Then the first number is 10x + y and the second number is 10y + x. We also have x + y = 9. So, x = 9 – y and 10x + y – (10y + x) = 27. Substituting for x, we get 10(9 – y) + y – (10y + 9 – y) = 27 90 – 10y + y – 10y – 9 + y = 27 81 – 18y = 27 18y = 54 y = 3. That means x = 9 – y = 9 − 3 = 6. The product of the two digits is 6 × 3 = 18 Ans.


The teaching tools of math counts practice problems and answers​ are guaranteed to be the most complete and intuitive. It helps students get inspired to explore and discover many creative ideas from themselves. and intuitive. It helps students get inspired to explore and discover many creative ideas from themselves.


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Level 3 2001

  • MP #3.1.1
  • MP #3.1.2
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Level 3 2002

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Level 3 2003

  • MP #3.3.1
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Level 3 2004

  • MP #3.4.1
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Level 3 2005

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Level 3 2006

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Level 3 2007

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Level 3 2008

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Level 3 2009

  • MP #3.9.1
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Level 3 2010

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Level 3 2011

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Level 3 2012

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Level 3 2013

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Level 3 2014

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Level 3 2015

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Level 3 2016

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Level 3 2017

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Level 3 2018

  • MP #3.18.1
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Level 3 2019

  • MP #3.19.1
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Level 3 2020

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  • MP #3.20.4