 Problem of the Week

Problem C and Solution

This Product is a Mystery

Problem

The number $$A8$$ is a two-digit number with tens digit $$A$$ and units (ones) digit 8. Similarly, $$3B$$ is a two-digit number with tens digit 3 and units digit $$B$$.
When $$A8$$ is multiplied by $$3B$$, the result is the four-digit number $$C730$$. That is,  If $$A$$, $$B$$, and $$C$$ are each different digits from $$0$$ to $$9$$, determine the values of $$A$$, $$B$$, and $$C$$.

Solution

In a multiplication question there are three parts: the $$multiplier$$, $$multiplicand$$ and $$product$$. In our problem, $$A8$$ is the multiplier, $$3B$$ is the multiplicand, and $$C730$$ is the product.
The units digit of the product $$C730$$ is 0. The units digit of a product is equal to the units digit of the result obtained by multiplying the units digits of the multiplier and multiplicand.
So $$8 \times B$$ must equal a number with units digit 0. The only choices for $$B$$ are 0 and 5, since no other single digit multiplied by 8 produces a number ending in zero.
However, if $$B=0$$, the units digit of the product is 0 and the remaining three digits of the product, $$C73$$, are produced by multiplying $$3\times A8$$. But $$3 \times A8$$ produces a number ending in 4, not 3 as required. Therefore $$B \not = 0$$ and $$B$$ must equal 5. So the multiplicand is $$35$$.
Since $$A8$$ is a two-digit number, the largest possible value for $$A$$ is $$9$$. Since $$98\times 35=3430$$, the largest possible value of $$C$$ is $$3$$. Also, the product $$C730$$ is a four-digit number so $$C\not = 0$$. Therefore, the only possible values for $$C$$ are $$1,2,\mbox{ and }3$$. We will examine each possibility for $$C$$.
If $$C=1$$, then $$C730$$ becomes $$1730$$. We want $$A8\times 35=1730$$. Alternatively, we want $$1730\div 35=A8$$. But $$1730\div 35\doteq 49.4$$, which is not a whole number. Therefore, $$C\not =1$$.
If $$C=2$$, then $$C730$$ becomes $$2730$$. We want $$A8\times 35=2730$$. Alternatively, we want $$2730\div 35=A8$$. Since $$2730\div 35=78$$, which is a whole number. Therefore, $$C=2$$ produces a valid value for $$A$$, namely $$A=7$$.
If $$C=3$$, then $$C730$$ becomes $$3730$$. We want $$A8\times 35=3730$$. Alternatively, we want $$3730\div 35=A8$$. But $$3730\div 35\doteq 106.6$$, which is not a whole number. Therefore, $$C\not =3$$.
We have examined every valid possibility for $$C$$ and found only one solution. Therefore, $$A=7$$, $$B=5$$ and $$C=2$$ is the only valid solution. We can easily verify that $$78\times 35=2730$$.