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CEMC Gauss (Grade 8), 2017

Algebra Difficulty 2.8 Junior Find the answer Canada

The values $2,3,4,$ and 5 are each assigned to exactly one of the letters $V, W, X$ , and $Y$ to give $Y^{X} -W^{V}$ the greatest possible value. The value of $X + V$ is equal to

Pick one

Solution

The units digit of the product $1ABCDE\times3$ is 1, and so the units digit of $E\times3$ must equal 1.

Therefore, the only possible value of $E$ is 7.

Substituting $E=7$ , we get

Since $7\times3=21$ , 2 is carried to the tens column.

Thus, the units digit of $D\times3+2$ is 7, and so the units digit of $D\times3$ is 5.

Therefore, the only possible value of $D$ is 5.

Substituting $D=5$ , we get

Since $5\times3=15$ , 1 is carried to the hundreds column.

Thus, the units digit of $C\times3+1$ is 5, and so the units digit of $C\times3$ is 4.

Therefore, the only possible value of $C$ is 8.

Substituting $C=8$ , we get

Since $8\times3=24$ , 2 is carried to the thousands column.

Thus, the units digit of $B\times3+2$ is 8, and so the units digit of $B\times3$ is 6.

Therefore, the only possible value of $B$ is 2.

Substituting $B=2$ , we get

Since $2\times3=6$ , there is no carry to the ten thousands column.

Thus, the units digit of $A\times3$ is 2.

Therefore, the only possible value of $A$ is 4.

Substituting $A=4$ , we get

Checking, we see that the product is correct and so $A+B+C+D+E=4+2+8+5+7=26$ .

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