I was just going to say, when I was interrupted, that one of the many ways of classifying minds is under the heads of arithmetical and algebraical intellects. All economical and practical wisdom is an extension of the following arithmetical formula: 2 + 2 = 4. Every philosophical proposition has the more general character of the expression a + b = c. We are mere operatives, empirics, and egotists until we learn to think in letters instead of figures.
Oliver Wendell HOLMES
Welcome to the blog Math1089 – Mathematics for All.
I’m glad you came by. I wanted to let you know I appreciate your spending time here on the blog very much. I do appreciate your taking time out of your busy schedule to check out Math1089!
Numbers are the backbone of mathematics. When these numbers are allowed to combine with various mathematical operations, several marvellous things can happen, like producing 1000 using eight 8’s.
In this blog post, we will solve this extremely important problem and it goes like this: Can you combine eight 8’s with any other mathematical symbols except numbers to represent exactly 1000? We may use basic mathematical symbols like addition, subtraction, multiplication, division, exponentiation, square root, etc. for our purposes. The simplest one goes like this:
888 + 88 + 8 + 8 + 8 = 1000.
Why? Why did we choose the numbers in such a way? What was the underlying idea behind choosing these numbers? Let’s analyze.
To make 1000 using eight 8’s, one’s digit has to add up to something ending in a zero. Since there are eight 8’s, there should be five numbers involved in the sum, so that we get the sum of 40 and hence, the one’s digit 0.
When we add the one’s digits to 40, we’ll need to carry the 4 into the ten’s place. The ten’s place will also need to add to something ending in a zero to make 1000. With carrying the 4, we’ll need two 8’s to make 16.
Now we’ll need to carry a 2; add an 8 to that, and we’ll get 10, which looks a lot like the start of 1000. Let’s see where we are: we need five numbers with a one’s digit, two with a ten’s digit, and one with a hundred’s digit. This gives us 888 + 88 + 8 + 8 + 8, or 1000.
Another way to look at it is by going back to the first assertion that there must be five terms. That leaves three eights to place, and the only possible combinations are:
8888 + 8 + 8 + 8 + 8 = 8920 (place all remaining 8’s in the first term);
888 + 88 + 8 + 8 + 8 = 1000 (place all remaining 8’s in the first and second terms);
88 + 88 + 88 + 8 + 8 = 280 (place all remaining 8’s in the first, second and third terms).
Next, we consider the products 8 × (8 + 8) = 128 and 128 × 8 = 1024. Hence, another representation is
8 × [(8 × 8) + (8 × 8)] – 8 – 8 – 8 = 1000.
Since 888 ⎼ 8 = 880, hence the other representation could be
(888 – 8) + 8 × (8 + 8) – 8 = 1000.
Again, with the value of 8888 ⎼ 888 = 8000, another way to represent it is
If we allow decimal numbers to come in, the following could be another representation:
Since 103 = 1000, using the exponentiation operation we can write
Allowing for the square root, we get √(8 + 8) = 4. Another solution could be
We know that 8/.8 = 10 and 103 = 1000. Another representation could be
As a product of two integers, we have the following solutions:
Few more representations are
Looking into various combinations, we can have the following solutions:
Another set of relations:
Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call “Math1089 – Mathematics for All!“.
Math1089 