Using the digits 2, 4, 5, 8, and 0 fill the boxes to get the:

(a) maximum product

(b) minimum product

(c) product greater than 150

(d) product nearest to 100

(e) product nearest to 5

Question

Using the digits 2, 4, 5, 8, and 0 fill the boxes to get the: (a) maximum product (b) minimum product (c) product greater than 150 (d) product nearest to 100 (e) product nearest to 5

Homi · Accepted Answer

We need to place the digits 0, 2, 4, 5, and 8 into the boxes. The boxes form two decimal numbers. The first number has one digit before the decimal point and two digits after it. Let us call this number $N_1 = A.BC$ . The second number has one digit before the decimal point and one digit after it. Let us call this number $N_2 = D.E$ . We need to find the product $N_1 \times N_2$ .

Step 1 — Maximum product

We want to make the product as large as possible. We should make the integer parts ( $A$ and $D$ ) as large as possible. The largest digits are 8 and 5. Let us try $A=8$ and $D=5$ . The remaining digits are 0, 2, and 4. To make $N_1 = 8.BC$ largest, we use $B=4$ and $C=2$ . So $N_1 = 8.42$ . To make $N_2 = 5.E$ largest, we use $E=0$ . So $N_2 = 5.0$ . Let us calculate the product.

$8.42 \times 5.0$

$= 42.10$

Now, let us try $A=5$ and $D=8$ . The remaining digits are 0, 2, and 4. To make $N_1 = 5.BC$ largest, we use $B=4$ and $C=2$ . So $N_1 = 5.42$ . To make $N_2 = 8.E$ largest, we use $E=0$ . So $N_2 = 8.0$ . Let us calculate the product.

$5.42 \times 8.0$

$= 43.36$

Comparing $42.10$ and $43.36$ , the maximum product is 43.36.

$\boxed{\text{Maximum product} = 43.36}$

Diagram 1

Step 2 — Minimum product

We want to make the product as small as possible. To get a very small product, we should use 0 as an integer part. Let us consider $N_1 = 0.BC$ or $N_2 = 0.E$ .

Case 1: Let $N_1 = 0.BC$ . The digit for $A$ is 0. The remaining digits are 2, 4, 5, and 8. To make $N_1 = 0.BC$ smallest, we use $B=2$ and $C=4$ . So $N_1 = 0.24$ . The remaining digits for $N_2 = D.E$ are 5 and 8. To make $N_2 = D.E$ smallest, we use $D=5$ and $E=8$ . So $N_2 = 5.8$ . Let us calculate the product.

$0.24 \times 5.8$

$= 1.392$

Case 2: Let $N_2 = 0.E$ . The digit for $D$ is 0. The remaining digits are 2, 4, 5, and 8. To make $N_2 = 0.E$ smallest, we use $E=2$ . So $N_2 = 0.2$ . The remaining digits for $N_1 = A.BC$ are 4, 5, and 8. To make $N_1 = A.BC$ smallest, we use $A=4$ , $B=5$ , and $C=8$ . So $N_1 = 4.58$ . Let us calculate the product.

$4.58 \times 0.2$

$= 0.916$

Comparing $1.392$ and $0.916$ , the minimum product is 0.916.

$\boxed{\text{Minimum product} = 0.916}$

Diagram 2

Step 3 — Product greater than 150

We found the maximum possible product in Step 1. The maximum product is 43.36. This value is not greater than 150. So, it is not possible to get a product greater than 150.

$\boxed{\text{Not possible}}$

Step 4 — Product nearest to 100

We found the maximum possible product in Step 1. The maximum product is 43.36. All possible products will be less than or equal to 43.36. Therefore, all products are less than 100. The product closest to 100 will be the largest possible product. This is 43.36.

$\boxed{\text{Product nearest to 100} = 43.36}$

Step 5 — Product nearest to 5

We need to find a product that is closest to 5. We will try different combinations of digits. Let us try to make one number $0.BC$ and the other $D.E$ . We want $0.BC \times D.E$ to be close to 5. Let us try to make $D.E$ large and $0.BC$ around $5/D.E$ .

Consider $D=8$ and $E=2$ . So $N_2 = 8.2$ . The remaining digits for $N_1 = 0.BC$ are 0, 4, and 5. We want $0.BC \approx 5 / 8.2 \approx 0.609$ . Let us try $B=5$ and $C=4$ . So $N_1 = 0.54$ . Let us calculate the product.

$0.54 \times 8.2$

$= 4.428$

The difference from 5 is $5 - 4.428 = 0.572$ .

Consider $D=5$ and $E=4$ . So $N_2 = 5.4$ . The remaining digits for $N_1 = 0.BC$ are 0, 2, and 8. We want $0.BC \approx 5 / 5.4 \approx 0.925$ . Let us try $B=8$ and $C=2$ . So $N_1 = 0.82$ . Let us calculate the product.

$0.82 \times 5.4$

$= 4.428$

The difference from 5 is $5 - 4.428 = 0.572$ .

Let us check other combinations to see if we can get a smaller difference. For example, let $N_1 = 0.52$ and $N_2 = 8.4$ . The digits used are 0, 5, 2, 8, 4. Let us calculate the product.

$0.52 \times 8.4$

$= 4.368$

The difference from 5 is $5 - 4.368 = 0.632$ . This is larger than $0.572$ .

Let us try to get a product slightly above 5. Consider $N_1 = 5.40$ and $N_2 = 0.8$ . The digits used are 5, 4, 0, 0, 8. This uses 0 twice. This is not allowed. We must use each digit exactly once. So $N_1 = 5.40$ (using 5, 4, 0). $N_2 = 0.8$ (using 8). The digits used are 0, 4, 5, 8. We are missing digit 2. This is not allowed.

We must use all five digits: 0, 2, 4, 5, 8. Let us recheck $N_1 = 5.42$ and $N_2 = 0.8$ . The digits used are 5, 4, 2, 0, 8. All digits are used. Let us calculate the product.

$5.42 \times 0.8$

$= 4.336$

The difference from 5 is $5 - 4.336 = 0.664$ . This is larger than $0.572$ .

The smallest difference we found is $0.572$ . This means the product 4.428 is nearest to 5.

$\boxed{\text{Product nearest to 5} = 4.428}$

Answer

(a) The maximum product is 43.36. (b) The minimum product is 0.916. (c) It is not possible to get a product greater than 150. (d) The product nearest to 100 is 43.36. (e) The product nearest to 5 is 4.428.

Question 34

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