# Do you know the answer to this problem 1/5×11_=51/_?

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• (1/5)*(110+x)=51/x

110+x=255/x

110x+x^2=255

x^2+110x-255=0

Now use the quadratic formula to solve for x

• If each underline is supposed to be a digit, let's set this up like an equation then.

The blank on the right can just be a letter "x".

The blank on the left is part of a larger number.  Since it's 11_, I'll call this 110 + y so when we know the value of y, the value works out.

So now we have:

(1/5)(110 + y) = 51/x

Let's start by multiplying both sides by 5x to get rid of the fractions:

x(110 + y) = 255

Now let's solve this for y in terms of x:

110x + xy = 255

xy = 255 - 110x

y = (255/x) - 110

We know that x and y must be integers between 1-9 (y can be 0 as well)

For y to be an integer, (255/x) must also be an integer We can look at the prime factorization of 255 to see what values are possible for x:

255 = 3 * 5 * 17

Since x must be less than 10, the only factors that qualify here are 1, 3, and 5.  Any other combination of factors makes this too large.

Now that we have two possible values for x we can see what the resulting ys are:

y = (255/x) - 110

y = (255/1) - 110 and y = (255/3) - 110 and y = (255/5) - 110

y = 255 - 110 and y = 85 - 110 and y = 51 - 110

y = 145 and y = -25 and y = -59

None of them fit as we need y to be an integer less than 10.

So either I misunderstood the question, or something else is wrong with it.