# Sunday Times Teaser 3011 – Optical Illusion

### by Danny Roth

#### Published Sunday June 07 2020 (link)

George and Martha are studying optics. If you place an object a specific distance from a lens, an image will appear at a distance from that lens according the following formula:

The reciprocal of the object distance plus the reciprocal of the image distance is equal to the reciprocal of the focal length of the lens.

The object distance was a two-digit whole number of cm (ab). The image distance and the focal length of the lens were also two-digit whole numbers (cd and ef respectively). The six digits were all different and non-zero. Also, the object distance and the focal length were of the same parity and b was an exact multiple of d. Martha pointed out that the sum of those three two-digit numbers was a prime number.

What was that prime number?

1. 2. 3. The lens equation 1/U+1/V=1/F has general solution in positive integers U=Am(m+n), V=An(m+n), F=Amn where A,m,n are integers and hcf (m,n)=1.

Since U+V+F is prime A=1 and since U and F have the same parity, m is even and n is odd.

A very small number of values of (m,n) can give 2-digit numbers with non-zero digits and a few lines of analysis and a small amount of calculation (or, I assume, a very simple program) give the solution and show its uniqueness.

U+V is square if and only if A is square.

It is possible to find the solution without using the primality and parity tests but it would need a lot of calculation (or, I assume, a more complex program) to show uniqueness.

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