“Almost an Integer” Problems
Posted on
20
Sep 2022

“Almost an Integer” Problems

Unless you do the math as a career or a hobby, you probably prefer integers to non-integers. Whole numbers are easier for us to conceptualize. But a certain class of GMAT/Executive Assessment (EA) problems involves numbers that are almost integers. Generally, this nearest integer is 1, so the nearby numbers look like this:

0.99999

0.9995

1.001

1.000006

Whenever you see a number like one of these on a GMAT/EA problem, you should use powers of 10 to notate the difference between the value in question and the nearest integer.

Let’s start by expressing our first example number, 0.99999, as a difference from 1 without powers of 10. Then we’ll convert that difference to a power of 10.

0.99999 = 1 – 0.00001

When the number you’re working with is of the form 0.999 . . . , the difference from 1 is all zeros with a 1 at the end, and the number of digits after the decimal remains consistent. Here there are five nines after the decimal, so our difference from 1 has four zeros and then a 1 at the end for a total of five digits after the decimal.

Now to convert our difference from 1 to scientific notation. The number in question, 0.00001, is small, so the power of 10 will be negative. The rule is to simply use the negative version of the number of digits after the decimal. According to this rule, 0.00001 = 10-5. Therefore we have this:

0.99999 = 1 – 0.00001 = 1 – 10-5

Let’s try the next example from above: 0.9995. This time there are only four digits after the decimal, but the last one is a 5 instead of a 9. Again, let’s express this value as a difference from 1 without scientific notation first:

0.9995 = 1 – 0.0005

As before, the number of digits after the decimal must remain consistent. But instead of using all zeros and then a single 1, we use all zeros and then whatever digit sums to 10 with the final digit of the original number. When the final digit of the original number is a 9, as in the first example, we use a 1 (since 9 + 1 = 10). In this case, the final digit of the original number is 5, so we need to use another 5 (5 + 5 = 10) to finish off our difference from 1.

0.9995 = 1 – 0.0005 = 1 – 5*10-4

There are four digits after the decimal, so the exponent of the 10 is -4. The coefficient of 5 is applied to the 10-4 term because 0.0005 = 5 * 0.0001 = 5*10-4.

We can solidify this with a general rule for finding decimal differences from 1. The number of digits after the decimal must remain consistent, and the digits in each place must sum to 9, except for the final digits which sum to 10. Here’s an example:

0.8653 = 1 – 0.1347

Here are the sums of the tenths, hundredths, thousandths, and ten-thousandths digits:

8 + 1 = 9

6 + 3 = 9

5 + 4 = 9

3 + 7 = 10

Here are some numbers you can use for practice. Their differences from 1 are at the end of the article.

0.23468

0.9834

0.31479

0.34098

0.999357

0.00042

0.000257

This covers numbers slightly less than 1. Numbers slightly greater than 1, like the examples from before of 1.001 and 1.000006, are easier to work with because you can convert everything after the decimal directly to a power of 10 without having to find a difference from 1.

1.001 = 1 + 10-3

1.000006 = 1 + 6*10-6

With these skills in place, you’re ready to tackle some official problems.

(1.00001)(0.99999) – (1.00002)(0.99998) =

(A) 0

(B) 10-10

(C) 3(10-10)

(D) 10-5

(E) 3(10-5)

Let’s convert each of the four numbers in the problem:

1.00001 = 1 + 10-5

0.99999 = 1 – 10-5

1.00002 = 1 + 2*10-5

0.99998 = 1 – 2*10-5

And we have a lovely pattern emerging.

(1 + 10-5)(1 – 10-5) – (1 + 2*10-5)(1 – 2*10-5)

Now all that remains is to “foil” the expressions and then simplify:

(1 + 10-5)(1 – 10-5) – (1 + 2*10-5)(1 – 2*10-5)

(1 – 10-5 + 10-5 – 10-10) – (1 – 2*10-5 + 2*10-5 – 4*10-10)

Now all the 10-5 terms cancel:

(1 – 10-10) – (1 – 4*10-10)

Now the 1s cancel as well:

-(10-10) + 4(10-10)

3(10-10)

And the correct answer is C.

Here’s another:

1 – 0.00001 = 

(A) (1.01)(0.99)

(B) (1.11)(0.99)

(C) (1.001)(0.999)

(D) (1.111)(0.999)

(E) (1.0101)(0.0909)

This one is different because the first step has already been done for us. Instead of starting with 0.999999, the problem starts in “difference from 1” form. All we have to do is convert the difference to a power of 10:

0.000001 = 10-6

1 – 0.000001 = 1 – 10-6

Now what to make of the answer choices? After a quick scan, the only ones that look very friendly to a “1 +/- 10-x” form are A and C. Choice A can’t be right because the factors 1.01 and 0.99 contain a total of only four digits after their decimals, and the product we are looking for, 1 – 10-6 or 0.999999, has six digits after the decimal. In multiplication, the number of digits after the decimal in the product always matches the total number of digits after the decimals in the factors.

So answer choice C looks like the best candidate. Let’s convert it to “power of 10” form:

(1.001)(0.999) = (1 + 10-3)(1 – 10-3)

Now we can “foil” the expression and simplify:

(1 + 10-3)(1 – 10-3) = 1 + 103 – 103 – 10-6 = 1 – 10-6

And as we suspected, answer choice C turned out to be correct.

Now you’re ready to handle “almost integers” on GMAT/EA problems. Next time we’ll use powers of 10 to address problems that ask about zeros or nonzero digits.

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Solutions to drills:

0.23468 = 1 – 0.76532

0.9834 = 1 – 0.0166

0.31479 = 1 – 0.68521

0.34098 = 1 – 65912

0.999357 = 1 – 0.000643

0.00042 = 1 – 0.99958

0.000257 = 1 – 0.999743

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