starting with 1:

(2 + 4) x 3 + 1

4 x 5 – 3 + 2

2 + 3 x 4 + 5

and other ways using four consecutive digits:

starting with 3: 3 x 6 + 5 – 4

starting with 4: 5 x 6 – 4 – 7 or 4 x 5 + 6 – 7

starting with 5: (8 – 6) x 7 + 5

starting with 6: 9 x 8 ÷ 6 + 7 or 8 ÷ 6 x 9 + 7 (a tough one... to find)

(2 + 4) x 3 + 1

4 x 5 – 3 + 2

2 + 3 x 4 + 5

and other ways using four consecutive digits:

starting with 3: 3 x 6 + 5 – 4

starting with 4: 5 x 6 – 4 – 7 or 4 x 5 + 6 – 7

starting with 5: (8 – 6) x 7 + 5

starting with 6: 9 x 8 ÷ 6 + 7 or 8 ÷ 6 x 9 + 7 (a tough one... to find)

it’s interesting that from those starting with 1:

4 x 5 – 3 = 3 x 4 + 5

and that from those starting with 4:

5 x 6 – 4 = 4 x 5 + 6

explore a generalisation for these results

and try to prove it

for any three consecutive numbers . . .

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