Integer Equations

Number Theory Level 1

$\color{#3D99F6}{x},$ $\color{#D61F06}{y},$ and $\color{#20A900}{z}$ are integers, and $\color{#3D99F6}{x}\color{#D61F06}{y}\color{#20A900}{z}=10.$ What is the maximum possible value of $\color{#3D99F6}{x}+\color{#D61F06}{y}+\color{#20A900}{z}?$

The answer is 12.

5 solutions

Eli Ross Staff
Oct 2, 2015

It's fairly clear it's best if they are all positive. Since $10=2\cdot 5,$ the only possibilities for the unordered positive integers $\{x,y,z\}$ are $\{1,1,10\}$ and $\{1,2,5\},$ so the maximum sum is $1+1+10=12.$

but then x=y=1 , z= 10.........means the equation will be x.x.z=10 and 2x + z = max value ..........which does nt fit the question i guess

Rahul Pal - 5 years, 8 months ago

The trick is that the problem doesn't force x, y, and z to be DISTINCT integers. If it did, then 8 would be the correct answer. Since the problem doesn't have that restriction, the combination of {1,1,10} gives the maximum sum.

Adam Bachmann - 5 years, 8 months ago

Vijay Simha
Oct 4, 2015

We can arrive at the maximum, possible value:

What is the minimum possible value of x + y + z ? (Decimals are ok) ?

Use AM > GM inequality, x + y + z is greater than equal to 3*cuberoot(10) = 6.46

Sadasiva Panicker
Oct 4, 2015

1x1x10 = 10: 1+1+10=12

Putu Pradnyana Iswara
Oct 4, 2015

1 x 1 x 10 = 10 1 + 1 + 10 = 12

Avinash Mahech
Oct 3, 2015

xyz=10 => z=10/xy let, x+y+z=k put z =10/xy ,we have k=x+y+10/xy dk/dy=0+1-10/x*y^2 dk/dy=0
so , x=10/y^2 now, k=x+y+10/xy k=10/y^2 +y+y k=10/y^2+2y Maximum value will give only if y=1 and we get the value of k=12

Integer Equations

The answer is 12.

5 solutions

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