Problem Solving

lheiannie07 wrote:In the product 2^19*9^13*5^24, what is the digit in the 20th place to the left of the decimal point?

A. 0
B. 2
C. 4
D. 5
E. 9

What is the correct solution to this to get the right Option? How will i formulate the formula in this?

OA D

Hi lheiannie07,
Let's take a look at your problem.
It seems a very interesting question.

2^19*9^13*5^24
Let's rewrite 5^24 as 5^19. 5^5

= (2^19)*(9^13)*(5^19)*(5^5)

Write the bases with the same exponents 19 together.
= (2^19)*(5^19)*(9^13)*(5^5)

We can write (2^19)*(5^19) as (2*5)^19
= ((2*5)^19)*(9^13)*(5^5)
= (10^19)*(9^13)*(5^5)

The term (10^19) indicates that this product has 19 zeros.
We are asked to find the digit at the 20th place to the left of the decimal point.
We already found out that there are 19 zeros to the left of the decimal point, so we need to find out the digit that comes right before the zeros.

To find out the digit right before the 19 zeros we need to look at the last digit of the product (9^13)*(5^5).
The last digit of 5^5 is a 5
Because when 5 is multiplied by itself odd number of times, the last digit of the product is always 5.
For example 5 * 1= 5, 5 *3 = 15, 5*5 = 25 etc

The last digit of 9^13 is an odd number
Because when 9 is multiplied by itself odd number of times, the last digit of the product is always odd.
For example. 9*1 = 9, 9*3 = 27, 9*5=45 etc

Now the last digit of 5^5 is a 5 and the last digit of 9^13 is an odd number, therefore the the last digit of the product of 5 and an odd number is again 5.

Therefore, the digit right before the 19 zeros will be the 5.

Hence, Option D is correct.
Hope this helps.

I am available if you'd like any follow up.

It seems suspicious that we'd be asked for the 20th place, so let's see if there's some reason we'd get that easily.

Noticing that we've got 2¹�, we might think of 10¹�, which would get us to the 20th place pretty nicely. (If it's not clear why, see my post below.) And what do you know, there it is! =>

2¹� * 5²� =>

2¹� * 5¹� * 5� =>

10¹� * 5�

Now let's bring in our 9¹³:

10¹� * 5� * 9¹³

So our 20th digit will be the units digit of 5� * 9¹³, which is just 5. (5 * anything odd ends in 5.)

If it's not clear why 10¹� would determine the 20th place in our number above, take a look at few smaller examples.

Suppose we have 10² * 5. That'll give us 500, and the 5 becomes the THIRD digit from the right.

Suppose we have 10� * 17. That'll give us 170,000, and the 7 (or the units digit of 17) is now the FIFTH digit from the right.

See the pattern? If we multiply by 10�, we shift everything n digits to the left, so the units digit of whatever we multiplied by will now be the (n + 1)th number from the right. Pretty cool!

2^19*9^13*5^24 = 2^19*9^13*5^19*5^5 = (2*5)^19*9^13*5^5 = 9^13*5^5*10000000000000000000

As any odd multiple of 5 has 5 as its last digit, then the 20th digit is 5.

If we can pair up multiples of 5 and 2, then each pair is a product of 10. The number of pairings is 19, so the last 19 places are zero.
The remaining factors are an odd number of 5s, resulting in the 20th digit being 5.