More Dicey Operations

The students at St Margaret's School for Girls in Scotland thought hard about this problem. Alexa began by looking for numbers that would multiply to 100:

I started to think about what numbers that would multiply to equal 100, like 50×2=100 and I put that into the question and tried to get really close to 100.

Evangeline thought about where to put larger and smaller numbers:

First I would see which numbers you can multiply by each other. If it was big numbers I was timesing, I would put the smaller number at the bottom and depending on what numbers were there I would put either the bigger/smaller number first, above the small number.

Syeda from Hollywood Primary in Australia used a similar strategy, but chose to use a large single digit multiplier when the top number was small:

When your top number is close to 100 try to multiply it by a small single digit number. Then when your top double digit number is low try to multiply it by a high single digit number. With this strategy your answer is likely to be correct.

Natalie from Lower Farm Academy in England looked for the best strategy when all three digits are small:

21×3 is the closest to 100 because you need to get the largest number as all of the digits are small. If all of the digits are small then you need to times a two digits number by the largest digit and the tens digit needs to be the second largest digit.

I wonder why the largest digit should be used as the single digit multiplier in this case?

Ayah from Doha College in Qatar tried this problem a lot of times. Here are two of the games Ayah played:

The numbers I received were 5, 2, and 4. I immediately recognized 25 × 4, since $\frac{1}{4}= 0.25$, $\frac{4}{4}= 1$ (or, in this case, 100), the target. Therefore, I did 25 × 4, which equals 100.

The numbers that were given to me were 3, 6, and 1. First, I eliminated 1 as the multiplier, since anything times 1 is itself, and 63 or 36 wasn't close to 100. Then I tried 16 × 3, but that gave me 48, a number too small. I then did 61 × 3 to get a high number that will potentially be closer to 100, but I got 183, which is far too much. So then I tried making 6 the multiplier and did 13 × 6, which gave me 78, so I tried that, since it only had a difference of 22.

Hank from Stowe School in the UK didn't think about numbers which multiplied to 100, but instead looked for numbers which multiplied to 10:

1. First find the pair of numbers that has the closest product to 10.  
2. If the product is smaller than 10, try both scenarios when you fill these two numbers in the tens digit of the top number and the ones digit of the bottom number.
3. If the product is greater than 10, fill the larger number in the tens digit of the top number and the smaller number in the ones digit of the bottom number.
4. Find the combination that has the product closest to 100.

Hank also found strategies for the two division versions of this game:

Div 1

1. Find the pair of numbers that are closest to each other.
2. If there are two digits that are equal, place one in the divisor, and the other in tens digit of the dividend.
3. If there are no equal digits, but two are consecutive numbers and both are not zero, place the larger number on the divisor and the smaller number on the tens digit of the dividend.
4. If none of the previous two scenarios is possible. Try other pairs of numbers that are close to each other and not 0. And place the larger number on the divisor, and the smaller number on the tenth digit of the dividend.
5. If there is one 0, you can only make it part of the dividend.
6. If there are two 0s, the only possible best solution for this case is 0.
7. Find the best possible scenario out of the scenarios described above.

Div 2

1. This case is almost the same as Div 1. Follow the same algorithm when selecting the divisor and the hundreds digit of the dividend.
2. When selecting the tens digit and the ones digit of the dividend, it depends on whether (hundreds digit × 100) / divisor currently is smaller or greater than 100. If it is smaller than 100, a two digit number that is the largest (made with the rest of the numbers) is preferable to become the tens and ones digits of the dividend. When it is greater than 100, a two digit number that is the smallest (made with the rest of the numbers) is preferable to become the tens and ones digits of the dividend.
3. Find the best possible scenario of the scenarios described above.

Arachyne from Donvale Christian College in Australia thought hard about the multiplication versions of this game, and worked on creating a general rule for how to win no matter how big the numbers are:

To ensure that you get as close to the target number as possible, you should only worry about the first digit of both numbers you are multiplying. For example, when trying to get to 100 with a two digit number and a one digit number, we should only worry about the one digit number and the tens place of the two digit number. Out of the three numbers that you pick, which two get you the closest to 100 / 10^1? 

This formula is T / 10^D - 2, which is T (the target number) divided by 10 to the power of D - 2 (the total number of digits in the numbers you're multiplying minus 2). 

Say, you got the numbers 5, 2 and 8. When you get a 5 and 2, those two will always be the first digits of both numbers you multiply, and this is because 5 × 2 = 10, which is the target value of T / 10^D - 2.

For this case, put the lower number in the front of the one digit number. This is because you multiply the one digit number by the ones place in the two digit number and you want that extra value to be as little as possible. So the multiplication problem would be 58 × 2. This will be how you arrange your numbers when the product of the two digits you plan to put on the front of the numbers you are multiplying is more than or equal to 10 (T / 10^D - 2 for other cases).

So, 5 and 2 would be the best solution, since it multiplies exactly to your target number. The next best solutions are here:
3 and 3 (9)
9 and 1 (9)
2 and 4 (8)
8 and 1 (8)
3 and 4 (12)

You're essentially trying to find two numbers that will get you closest to the number 10. In the case where the product of the two digits you plan to put on the front of the numbers you are multiplying is less than 10, you want to try to get more value in your answer. This means essentially, you're going to do the opposite of what should be done if the product of both digits is more than or equal to 10. The further your number is from the answer, the more value you will want to earn.

This will not always work, so you should try rearranging the last few digits to get as close as possible. Move from left to right, the closer to the left the digits are, the higher value they have, so moving them will have a large effect. Mostly, this problem is a matter of trial and error, check your answer and see if you can get closer. It is more likely to work when the numbers are further away from the target numbers.

We also received similar solutions from: Kazuto and Hodaka from St. Mary's International School in Japan; Ziqian from Harrow International School in Hong Kong; Krishanu from India; Sienna from Copthorne Prep School in the UK; Vihaan from Singapore American School; Amelia from Doha College in Qatar; Nakul and Rudranil from The GYM Foundation in Finland; and Avic from Ganit Kreeda in Vichar Vatika, India. Thank you all for sharing your ideas with us.