SOLUTION: Use the clue to complete the equation using some or all of the digits 2, 4, and 6. Clue: The product is between 2,600 and 2,700. ?×?5?=26??

?×?5?=26??
A×B5C=26DE

That gives the equation

A(100B+50+C)=2600 + 10D + E

{A,C} is not equal to {2,4} since that would make E=8.

So we have only 4 cases to consider:

Case 1: A and C, respectively either equals 2 and 6, respectively, and E=2 

Case 2: A and C, respectively either equals 6 and 2, respectively, and E=2 

Case 3: A and C, respectively either equals 4 and 6, respectively, and E=4 

Case 4: A and C, respectively either equals 6 and 4, respectively, and E=4 

A and C, respectively equals 2 and 4 and E=4

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Case 1:  (A,C) equals (4,6) and E=2 

2×B56=26DE

4(100B+50+6)=2602 + 10D  

2(100B+56)=2602 + 10D

200B+112=2602 + 10D
200B = 2490 + 10D

20B = 249 + 10D
20B-10D = 249
The left side is divisible by 10 but the 
right side is not. 

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Case 2:  {A,C} equals {6,2} and E=2 

6×B52=26D2

6(100B+50+2)=26D2  or 6(100B+50+2)=26D2

6(100B+52)=2602 + 10D

200B+312=2602 + 10D
200B = 2290 + 10D

20B = 229 + D
The left side is divisible by 20 but the 
right side cannot be unless D could be 11.

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Case 3:  (A,C) equals (6,4} and E=2 

6×B52=26D2

6(100B+50+2)=26D2  or 6(100B+50+2)=26D2

6(100B+52)=2602 + 10D

600B+312=2602 + 10D
600B = 2290 + 10D

60B = 229 + D

The left side is divisible by 60 but the 
right side cannot be unless D could be 11,
which it cannot.

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Case 4:  (A,C) equals (4,6) and E=4 
4×B56=26D4

4(100B+50+6)=26D4

4(100B+50+6)=26D4  or 6(100B+50+4)=26D4

4(100B+56)=2604 + 10D

400B+224=2604 + 10D
400B = 2380 + 10D
40B = 238 + D
D = 40B - 238

The right side is divisible by 2 but not by 4, so D=2
2 = 40B - 238
240 = 40B
  6 = B

So the solution is 4×656=2624.

Edwin