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In the multiplication at the right, each letter represents a different digit. If A is not zero, what are the values of A,B,C and D?

A B C

× C

Equal D B C

1
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A B C

× C

Equal D B C

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ABC x C = DBC

(100A + 10B + C) C = (100D + 10B + C)

100AC + 10BC + C² = 100D + 10B + C

Since they are equal, they must have the same units digit.**The units digit of C² must be equal to C. **

C 0 1 2 3 4 5 6 7 8 9

C² 0 1 4 9 6 5 6 9 4 1

C can be 0, 1, 5, or 6.

**100AC will have a maximum value of 900** since we have 100D at the right side of the equation and D would only be a digit (meaning it ranges from 0 to 9).

From this part it becomes a bit tricky.

100AC + 10BC + C² = 100D + 10B + C

*If C = 0*

0 = 100D + 10B

A digit cannot be equal to 0 so C≠0.

*If C = 1*

100A + 10B + 1 = 100D + 10B + 1

100A = 100D

Since two letters represent different digits C≠1.

*If C = 5*

500A + 50B + 25 = 100D + 10B + 5

500A - 100D + 40B + 20 = 0

25A - 5D + 2B + 1 = 0

5(5A - D) = - (2B + 1)

Since**B is positive **this would mean that **D>5A**

A = 1 since A cannot be 0 and D ranges from 0 to 9

5(5 - D) = -2B - 1

25 - 5D = -2B - 1

26 = 5D - 2B

Since 26 and -2B are even, D is even. Since D is greater than 5 the only possible values of D would be 6 and 8. The corresponding values of B would be 2 and 7.

A = 1, B = 2, C = 5, D = 6

A = 1, B = 7, C = 5, D = 8

*If C = 6*

600A+ 60B + 36 = 100D + 10B + 6

600A - 100D + 50B + 30 = 0

60A - 10D + 5B + 3 = 0

5(12A - 2D + B) = -3

Since they do not have equal signs, 12A - 2D + B is negative meaning**12A + B < 2D**. This would mean A=1 (**because A≠0**), we cannot have a higher value of A since that would make 12A greater than 18 which means that 2D cannot be higher than 12A + B (since D ranges from 0 to 9).

5(12 - 2D + B) = -3

60 - 10D + 5B = -3

63 = 10D - 5B

63 = 5 (2D - B)

63 is not a multiple of 5 so C≠6.

*Final Answer***:**

A = 1, B = 2, C = 5, D = 6

A = 1, B = 7, C = 5, D = 8

(100A + 10B + C) C = (100D + 10B + C)

100AC + 10BC + C² = 100D + 10B + C

Since they are equal, they must have the same units digit.

C 0 1 2 3 4 5 6 7 8 9

C² 0 1 4 9 6 5 6 9 4 1

C can be 0, 1, 5, or 6.

From this part it becomes a bit tricky.

100AC + 10BC + C² = 100D + 10B + C

0 = 100D + 10B

A digit cannot be equal to 0 so C≠0.

100A + 10B + 1 = 100D + 10B + 1

100A = 100D

Since two letters represent different digits C≠1.

500A + 50B + 25 = 100D + 10B + 5

500A - 100D + 40B + 20 = 0

25A - 5D + 2B + 1 = 0

5(5A - D) = - (2B + 1)

Since

A = 1 since A cannot be 0 and D ranges from 0 to 9

5(5 - D) = -2B - 1

25 - 5D = -2B - 1

26 = 5D - 2B

Since 26 and -2B are even, D is even. Since D is greater than 5 the only possible values of D would be 6 and 8. The corresponding values of B would be 2 and 7.

A = 1, B = 2, C = 5, D = 6

A = 1, B = 7, C = 5, D = 8

600A - 100D + 50B + 30 = 0

60A - 10D + 5B + 3 = 0

5(12A - 2D + B) = -3

Since they do not have equal signs, 12A - 2D + B is negative meaning

5(12 - 2D + B) = -3

60 - 10D + 5B = -3

63 = 10D - 5B

63 = 5 (2D - B)

63 is not a multiple of 5 so C≠6.

A = 1, B = 2, C = 5, D = 6

A = 1, B = 7, C = 5, D = 8