# If 31x + 30y + 29z =366 , find x+y+z ? (x,y,z are natural numbers)

3,502 views
posted May 13, 2015

## 3 Solutions

Very tricky

366 days in a year -

31*7+30*4+29*1= 366

So x+y+z = 12 i.e. number of months in a year

solution May 13, 2015
There are two answers:  first set:  1, 4, 7
second set:  2, 2, 8
The above answers were arived at visually by using a simple matrix consisting of 3 columns (x,y,z) and 10 rows (29x1, 29x2 etc, then 30x1, etc, then 31x1 etc. and going up to 10 for each multiplication).  And just by looking at the numbers it should only take a few attempts to get the 2 sets of number.
+1 vote

Considering the number 29,30,31 as number of days of month and number of days in leap year - 366 .
= 29*1+ 30*4+31*7
= 366

solution Jun 20, 2016 by anonymous
+1 vote

Simple solution for 2 different answers using a matrix:

solution Sep 13, 2016

Similar Puzzles

Using algebra or matrix mechanics, solve for x, y, and z. Where x, y, and z are positive integers, and where zero is not an integer or part of an integer.

29x+30y+31z=366. What is the value of x , y, and z? In order to answer the question correctly, you must provide 2 sets of numbers which will satisfy the equation.

Hint: A simple leap year calendar will solve for one set.

Hint #2: Algebra will also solve for another/both sets of numbers. However, a simple 2D matrix (columns and rows) will allow for a quick and easy visualization of the 2 sets of numbers.

Conclusion: Matrices and Feynman type diagrams are often more powerful tools than algebra. Learn to use both, for they are very effective problem solving devices in situations where algebra is too tedious or complicated. And for those that are mathematically challenged and think in terms of pictures rather than numbers, Matrices and Diagrams may be the best solution.

+1 vote

x and y are distinct 2 digit numbers such that y is obtained by reversing the digits of x.
Suppose they also satisfy x^2 - y^2 = m^2 for some positive integer m, then find the value of x+y+m?

``````x^2+y-z = 42