EASY but the toughest one!!!

Algebra Level 3

If $x, y, z$ are positive integers such that $31x + 29y + 30z = 366$ , then find $x+y+z$ .

The answer is 12.

2 solutions

Noel Lo
Apr 7, 2015

Since x, y and z > 0, 29(x + y + z) < 31x + 29y + 30z < 31(x + y +z). It follows that 29(x + y + z) < 366 < 31(x + y +z) so that $\frac{366}{31} < x + y +z < \frac{366}{29}$ or 11.8 < x + y + z < 12.6. Since x, y and z are integers, the only possible value for x + y + z is $\boxed{12}$ .

Very nice solution...

Priyanshu Mishra - 6 years, 2 months ago

x+y+z represents the number of months in the year

Kushagra Sahni - 5 years, 7 months ago

Yeah, nice thinking. Thats what the problem is set for.

Priyanshu Mishra - 5 years, 7 months ago

U J
Mar 18, 2018

30(x+y+z)+x-y=366 now,both lhs and rhs would leave same remainder when divided by 30,as lhs=rhs dividing both side by 30 we get remainder x-y=6(as x-y is smaller than 30) this gives y=x-6........| plugging | in original equation we get 60x+30z=540 = 2x+z=18 it is easy to see that for +ve soln many pair of x and z are there each would lead to same sum(after finding y) i.e 12

Good I like it but I have something different..... Keep it up.
Good luck..

Ayush Kumar - 3 years, 2 months ago

You found $y=x-6$ and $z=18-2x$ . This means $x+y+z=x+x-6+18-2x=12$ . I'm not sure if you noticed that. But nice solution regardless.

James Wilson - 7 months, 1 week ago

EASY but the toughest one!!!

The answer is 12.

2 solutions

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