In the last math post, we dealt with exponents and really big numbers in quantitative comparison. Let’s now try the same but in the problem solving context.

The general approach is not too different—you want to look for a pattern. However this time you will have to be exact, instead of simply saying which side is bigger. One tactic the GRE employs is to make a problem look impossible to solve in 2 minutes. But don’t despair. All problems are solvable once you unlock them. So if your approach is taking too much time, step back and look at the problem from a new angle.

Try the following and see if you can solve it in fewer than 2 minutes.

What is the sum of the digits of integer x, where x = 4^10 x 5^13?

(A) 13

(B) 11

(C) 10

(D) 8

(E) 5

If you spotted the pattern early on, you may have only required a minute. But if you couldn’t unlock the problem, here is how you do so. Notice that 4^10 can be rewritten as 2^20. We can now express x as 2^20 x 5^13. The logic here is that 2 x 5 = 10. That is, 10 to any integer power greater than 1,will be a 1 followed by zeroes

So now let’s rewrite the problem again so we get 2^7 x 2^13 x 5^13. Combine 2^13 x 5^13 and we get 10^13. That is, we get 1 followed by 13 zeroes. If you are taking the sum, it’s straightforward: 1 plus 13 zeroes is 1. We are not done yet as we have the 2^7. When you multiply this out, you get 128. 128 x 1,000 thirteen zeroes is equal to 128 followed by the thirteen zeroes. Ignore the zeroes and we get 1 + 2 + 8, which equals 11. Answer B

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In the practice question Why power of y equals 17?

Shouldn’t it be y to 12-(+5)=7

Y has -5 in denominator. Will it not become +5 in Numerator?

Hi Azam,

I’m not sure if I understand this question–are you sure you wrote it on the correct blog post? The practice question in this blog post uses the variable ‘x’ and doesn’t have a fraction in it. I’d be happy to answer your question if you let me know where it came from!

can we solve this problem by unit digit?

4^10*5^13

unit digit of 4^10 is 6

unit digit of 5^13 is 5

6+5=11

Hi Aditya,

Unfortunately, focusing only on the units digit for this type of problem will not work. In this specific situation, the answer worked out by coincidence. The question asks us for the “sum of digits of integer x”. Technically, if “x = 4^10 x 5^13”, we should not be adding the units digit of “4^10” and “5^13”. Within “x” we are told those two numbers (therefore, the units digits) must be

multipliedtogether. As such, this approach is not valid, but I like your thinking!Hi Magooshers,

I need your help with my third problem please,

If I have a small circle in a large circle, both circles share a 90 degree sector. If no other information is given. How can I compare the circumference of the small circle to the length of Arc of the bigger circle?

Are the quantities equal, one is smaller, larger or can’t be decided based on information given?

Thank you

Hi Raina,

Because you are a premium member and your questions are rather intricate, I have sent them over for our team of test prep experts to help you with! You should be hearing from them soon. 🙂

Hi Magoosh Expert,

Thank you so much:)

Hey Chris,

I noticed that you can make this question slightly trickier by adding 12 as an answer choice.

I noticed that while solving the question I accidentally counted the 10^13 as 1(as 1 + 13 zeros = 1) and then counted 128 separately as 11. So I got a wrong answer of sum of digits = 12, but since 12 was not part of the answers I quickly realized my mistake.

Good question though 🙂

That’s true! Good eye :). I’ll make the change now.

How to solve something like

Find the greatest integer

a) 10^10 + 2^100

b) 100^10 + 2^10

c) (100 + 2 ) ^10

b) can be re written as 10^100 +2^10, which you can intuitively see that is bigger than a)

(because a. has the same powers but and the same bases but switched up)

So a<b

Then, looking at c) we see that it is definitely bigger than b) because,

using FOIL method, c can be written as 100 ^10 + 2^10 + other terms

So, b<c

Combining both results above we get a<b<c

Hi chris, I tried a crude method and had the same answer.

this is how my method worked:

4 raised to an even number will always yield 6 as its unit number and 5 raised to any number yields 5 as its unit number, add them up and you will have 11. Is it a perfect method?

Hmm..actually, I think that you got luck in this case :). The question is actually asking for the sum of all of the digits that make up ‘x’, hot the sum of the units digit of the two numbers. That said, I might have to rewrite this question, because I’m pretty sure you are not the first person who used this approach 🙂

I loved this question 🙂

Glad you like it 🙂

What a great Problem !!! I have to appreciate you !!! You are blessed with great Mind.

You’re so right!!!!

Thanks!

Hi there,

I got a slightly different answer and I am not being able to see what am I doing wrong…. (…. starting to think it could be the effect of studying too much!….)

I solved the problem as follows:

x=4^10 * 5^13

x= 4^10 * 5^10 * 5^3

x= 20^10 * 5^3 = 2 + 1+ 2 + 5 = 10 … not 11…

This one is tricky – at the very end you have 20^10. This is an astronomical number and therefore it is difficult to realize that the sum of it’s digits does not equal 2. If you break up 20^10 to 2^10 x 10^10 you get a number that has the first four digits of 1,024 (followed by a massive string of zeroes). If you try to multiply this number times 5^3 things get even messier.

So therefore, we want to simply this massive number at the very beginning by changing 4^10 to 2^20. Then we can share 13 of those zeroes with 5^13: 2^7 x (2^13 x 5^13), which give us 2^7 x 10^13. Now all we have to do is multiply out 2^7 getting 128. 10^13 gives us 13 zeroes, which we can discount when finding the sum of the digits. So we have 1+2+8 = 11.

This is a tough problem that only about 20% of test takers would be able to answer in less than 3 minutes.

The problem is that 20^10 is not 2 and ten zeroes (as if it was 10^10), it’s 1024 and ten zeroes.

But still, your method looks alright and the sum would be 1+2+4+1+2+5=15, not 11, which puzzles me…

Hi Albert,

The thing is if you just look at 20^10, then, yes, you get 1024 followed by 10 zeroes. But you have to multiply that result by the remaining 5^3. Don’t forget that 4^10 x 5^13 equals 20^10 x 5^3. Let me know if that makes sense :).

20^10 x 5^3 can be simplified to 2^10 x 10^10 x 5^3 right?

So that is 1024 x 10^10 x 5^3

Which is 1024 x 625 x 10^10

Which is 640000 x 10^10

So the answer is 6+4 = 10 right??

So where did I go wrong?

Akhil,

It looks like you made a slight mistake in the second line, where you translate 5^3 as 625.

Hope that helps!

Thanks Chris. Silly Me!

No prob 🙂