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Let's the value be m and its shifted version n.

2/3*n = m

m and n are both integers, so n-m=1/3*n is also an integer.

Therefore 3 | n .

As their digits have the same sum, 3 | m.

Therefore 3 | n-m --> 3 | 1/3*n --> 9 | n.

As their digits have the same sum, 9 | m.

Following similar logic from above, above 9 | n-m --> 9 | 1/3*n --> 27 | n.

So you need only account for 1/27 of the integers for a brute strength solution (which speeds it up significantly)

I got 285714 and 571428.

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Let's the value be m and its shifted version n.

2/3*n = m

m and n are both integers, so n-m=1/3*n is also an integer.

Therefore 3 | n .

As their digits have the same sum, 3 | m.

Therefore 3 | n-m --> 3 | 1/3*n --> 9 | n.

As their digits have the same sum, 9 | m.

Following similar logic from above, above 9 | n-m --> 9 | 1/3*n --> 27 | n.

So you need only account for 1/27 of the integers for a brute strength solution (which speeds it up significantly)

I got 285714 and 571428.

brilliant observation

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