# Beat The GMAT Challenge Question – August 16, 2010

by on August 16th, 2010

Congrats to Jason for being selected as this week’s winner of the BTG Challenge Question!  Jason will receive a Beat The GMAT t-shirt as a prize. Above you will find the full video explanation to the question below:

This week’s challenge question has no answer choices.  You’ll have to solve it all on your own.

Of those who respond with the correct answer, one person will be randomly selected to receive Beat The GMAT T-Shirt!  Deadline to enter is 5PM PDT, Tuesday, August 17, 2010. Good luck!

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How many positive divisors of 5,400,000 are perfect squares?

• IMO 17;

Breaking down into factors the number 54*10^5, we have (2^6)*(3^3)*(5^5)

2^6 will have 3 perfect squares: 2^2, 4^2, 8^2;
3^3 will have 1 perfect square: 3^2;
5^5 will have 2 perfect squares: 5^2, 25^2

Now each of these perfect squares :
(2^2, 4^2, 8^2),
(3^2),
(5^2, 25^2) will have mutual factors
example 2^2*3^2 etc.,

so, 2^2, 4^2, 8^2 will have 3+3+3 with (3^2), (5^2, 25^2)
and (3^2), (5^2, 25^2) will have 2 perfect squares.

so we get 3+1+2+9+2=17;

• The number can be written as 2^6 * 3^3 * 5^5 in prime factors.

Now we have three numers to pick from:
There could be three possibilities for 2 to be in the perfect square : 2^2, 2^4 and 2^6. So we have 3 ways.

For 3, there could be only one way : 3^3. So just 1 way
For 5, there could be two ways : 5^2 and 5^4. So two ways

Factors with all three primes: 3*1*2=6
Factors with any two primes : 3+2+6=11
1. 3(1 way) and 2(3 ways) would be 3(1*3) ways
2. 3(1 way) and 5(2 ways) would be 2(1*2) ways
3. 2(3 ways) and 5(2 ways) would be 6(3*2) ways

Factors with only one prime: 1+3+2=6
1. 3 - 1 way
2. 2 - 3 ways
3. 5 - 2 ways

Total : 6+11+6 = 23 positive divisors are perfect squares.

• Ans. 63

5,400,000 = 6 * 9 * 100000
= 6 * (3^2) * (10^5)
= (2*3) * (3^2) * (2*5)^5
= 2^6 * 3^3 * 5^5
= 2^6 * 3^2 * 3 * 5^4 * 5

2^6 , 3^2 and 5^4 are distinct prime factors in terms of squares
The squares that can be formed with each distinct prime factors 2,3,5 separately are:
2^2= 4
2^4 =16
2^6 = 64

3^2 = 9

5^2 = 25
5^4 = 625

Hence the total no. of squares that can be formed by the combination of these 6 squares - 4,16,64,9,25,625 are:

6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 63

• 2^2, 2^4,2^6,3^2,5^2,5^4

6! = 720

• Ans. 64

Oops forgot to include divisor 1 earlier as it's a perfect square

• Answer is 24. Explanation below.

The number 5400000 can be written as 2^6 * 3^3 * 5^5. We are looking for perfect squares, so we can write the number as 2^6 * 3^2 * 5^4 * 3 *5

after removing the solitary 3 and 5 - which can not contribute towards perfect square, the number is 2^6 * 3^2 * 5^4.

lets write it as 2^a * 3^b * 5^c.

In a perfect square :
a can take four values 0, 2,4, 6
b can take two values 0, 2
c can take three values 0, 2,4

so, number of combinations = 4 * 3 * 2 = 24

• Final Ans. 30 (Sorry, earlier I counted extra combinations)

5,400,000 = 6 * 9 * 100000
= 6 * (3^2) * (10^5)
= (2*3) * (3^2) * (2*5)^5
= 2^6 * 3^3 * 5^5
= 2^6 * 3^2 * 3 * 5^4 * 5

2^6 , 3^2 and 5^4 are distinct prime factors in terms of squares
The squares that can be formed with each distinct prime factors 2,3,5 separately are:
2^2= 4
2^4 =16
2^6 = 64

3^2 = 9

5^2 = 25
5^4 = 625

Hence the total no. of squares that can be formed by the combination of these 6 squares - 4,16,64,9,25,625 are

6C1 + 6C2 - 3 + 6C3 - 11 + 2
= 6 + 15 -3 + 20 - 11 + 2
= 6 + 12 + 9 + 2
= 29

(Note: This time the extra combinations that won't qualify are excluded. For e.g we deduct 3 from 6C2 as the following 3 pairs won't qualify
(4,64) (16,64) and (25,625). By the same reasoning 11 is deducted from 6C3. Only 2 combinations will qualify when we group by 4. None will qualify when we group by 5 or 6 )

Finally including divisor 1 as it's a perfect square too
Total divisors = 29 + 1 = 30

• There are the following prime factors of 5400000

3,3,3,2,2,2,2,2,2,5,5,5,5,5

If you find pairs of prime factors (because we are looking for perfect sqaure divisors), we have the following:

3,2,2,2,5,5

Thus we have 6 possible prime factors pairs. All the possible combinations of these factors will give us the number of possible divisors (factors) of 5400000 that are perfect squares.

The number of such factors are:

6C0+6C1+6C2+6C3+6C4+6C5+6C6 which is equal to 63.

Therefore there are 63 divisors of 5400000 that are perfect squares.

• I came up with 24:

5,400,000 = 2^6 x 3^3 x 5^5
Factors that matter = 2 ^6 x 3^2 x 5 ^4

1 set squares
2 has 3 sets of squares (2^2, 2^4, 2^6)
3 has 1 set of squares (3^2)
5 has 2 sets of squares (5^2, 5^4)

That is 6 sets of squares total

2 Set combinations:

Combinations of 2 and 3 = 3 x 1 = 3
Combinations of 2s and 5s = 3 x 2 = 6
Combinations of 5 and 3 = 2 x 1= 2
Total = 11 squares

3 Set combinations:

Combos of 2/3 = 3
Possibilities of 5s= 2
Total sets = 3 x 2 = 6

0 Set combinations=
2^0, 3^ 0, 5^0 = 1

Total = 6 + 11 + 6 + 1 = 24 perfect squares in 5,400,000

• Congrats to Jason for being selected as this week's winner!

Explainaition:

Factors of 5,400,000 : (2^6)*(3^3)*(5*5)

So the total Number of divisors of 5,400,000 are (6+1)*(3+1)*(5+1)=168

now out of 168 positive divisors we need to find out the number of divisors which are perfect squares
_____________________________________
we will take each prime factor one be one for number of perfect squares
1. (2^6)=(2^2), (2^4), (2^6) total =3
2. (3^3)=(3^2) total =1
3. (5^5)=(5^2), (5 ^4) total =2
_____________________________________
Now taking combination of two prime factors

1. (2^6)*(3^3)=considering the lowest power out of two we have [(2*3)^2] so total =1

2.(2^6)*(5^5)=again considering lowest power [(2*5)^5] this means we have two factors (2*5)^2 , and (2*5)^4 so total =2

3. (3^3)*(5^5)=here also lowest power (3*5)^3 mean we have one factor (3*5)^2 so total =1
____________________________________
now taking all three prime factor together

1. (2^6)*(3^3)*(5^5) here taking the lowest power we have (2*3*5)^3 so we have only one factor i.e (2*3*5)^2 total =1

********************************
taking the sum of all the no of perfect squares from above
i.e 3+1+2+1+2+1+1=11

THE CORRECT ONE IS this one

Explaination:

Factors of 5,400,000 : (2^6)*(3^3)*(5*5)

So the total Number of divisors of 5,400,000 are (6+1)*(3+1)*(5+1)=168

now out of 168 positive divisors we need to find out the number of divisors which are perfect squares
_____________________________________
we will take each prime factor one be one for number of perfect squares
1. (2^6)=(2^2), (2^4), (2^6) total =3
2. (3^3)=(3^2) total =1
3. (5^5)=(5^2), (5 ^4) total =2
_____________________________________
taking combination of two each at time

1. (2^6*3^3): so the perfect squares are (2^2*3^2),(2^4*3^2),(2^6*3^2) SO Total =3
2: (3^3*5^5): so the prefect squares are (3^2*5^2),(3^2*5^4)
so total =2
3: (2^6*5^5): so the perfect squares are (2^2*5^2),(2^4*5^2),(2^6*5^2),(2^2*5^4),(2^4*5^4),(2^6*5^4) so total=6

________________________________
taking combination of 3 together

(2^6*3^3*5^5): so the perfect squares are (2^2*3^2*5^2),(2^4*3^2*5^2),(2^6*3^2*5^2),(2^2*3^2*5^4),(2^4*3^2*5^4),
(2^6*3^2*5^4) so total=6

answer sum of all perfect squares comes out to be
3+1+2+3+2+6+6=23

• Sandy,

I think you are missing "1" (a perfect square). as one of the possible divisors here

Hence answer will 23 + 1 = 24

• The number can be written in perfect square form as:

4^3 * 9^1*25^2*3*5

Forget about the 3*5 since they dont form a perfect square, which leaves us:
4^3*9^1*25^2

The number of factors for a^m*b^q=(m+1)(q+1)

so in our case, (3+1)(1+1)(2+1)= 4*2*3=24

So 24 is the answer. Now give me the T-Shirt

• Yes, the ans. should be 30 - 6 = 24.

I counted extra 6 combinations that resulted from square pairs 4 and 16 as their product 4 * 16 = 64 is also one of the squares

• 2 2 2 2 2 2 3 3 3 5 5 5 5 5
2^6 3^3 5^5
4^3 9 25^2 3 5

1

2
2 2
2 2 3
2 2 3 5
2 2 3 5 5
2 2 5
2 2 5 5
2 2 2
2 2 2 3
2 2 2 3 5
2 2 2 3 5 5
2 2 2 5
2 2 2 5 5
2 3
2 3 5
2 3 5 5
2 5
2 5 5

3
3 5
3 5 5

5
5 5

= 24

• We have the following perfect square numbers:
2^2, 2^2, 3^2, 5^2, 5^2.

Now, we can select ONE of them in 3 ways (as 2^2 and 5^2 appear twice we dont need to double-count them).
We can select TWO of them together in 5 ways (such as 2^2 & 2^2 OR 3^2 & 5^2)
We can select THREE of them together in also 5 ways (such as 2^2 & 2^2 & 3^2 OR 2^2 & 3^2 & 5^2)
We can select FOUR of them together in 3 ways (such as 2^2 & 3^2 & 5^2 & 5^2 OR 2^2 & 2^2 & 5^2 & 5^2)
We can select all 5 of them in 1 way.

In total we have 3+5+5+3+1 = 17 ways of selecting perfect square divisors.

Ans 17.

Let me know what you think.

• Forgot to add 1 as a positive divisor. So the answer should be 17+1 = 18

• There's an easy way to crack this.
Replace 2^2 with x, 3^2 with y and 5^2 with z.

So, in terms of rewriting the number 5,400,000 with factors that are perfect squares, we have

5,400,000 = k*(x^2)*y*(z^2) where k accounts for the remaining divisors.

The number of divisors that are perfect squares are a combination of (x^2)*y*(z^2). Hence there are (2+1)*(1+1)*(2+1)=18 divisors that are perfect squares.

• 5,40,000 = 2^6 * 3^3 * 5^5
Now perfect square part 2^6 * 3^2 * 5^4
=
(1, 4, 16, 64) * (1, 9) * (1, 25, 625)

Total no of perfect square = 4*2*3 = 24

• 540000 = 2^6 * 3^3* 5^5

Number of divisible for a number is equal to (n+ 1) * (m+1) * (z+1) *....
where n,m, z... are the power to which the prime factor of the number is raised to.

Now coming back to the question, 540000 = 2^6 * 3^2 * 5^4 * 5*3
Since any number to be the perfect square as well as be divisor of 540000 should be able to be represented by (2^2)^3 or (3^2)^1 or(5^2)^2 or combination of these
hence the answer will be (3 + 1)* ( 1+ 1) * ( 2 + 1) = 4*2*3 = 24

• How many positive divisors of 5,400,000 are perfect squares?

Let's say,
N = 5,400,000
= 2^6 * 3^3 * 5^5
= (2^6 * 3^2 * 5^4) * 3 * 5

For 2^6, number of factors that are perfect squares = 4
(i.e. 1, 4, 16, 64)

For 3^2, number of factors that are perfect squares = 2
(i.e. 1, 9)

For 5^4, number of factors that are perfect squares = 3
(i.e. 1, 25, 625)

Therefore, the number of positive divisors of 5,400,000 that are perfect squares = 4 * 2 * 3
= 24

-------------------------------------------------------------------

• I beleive all of us missing small thing in the end :

N = 5,400,000
= 2^6 * 3^3 * 5^5
= (2^6 * 3^2 * 5^4) * 3 * 5

For 2^6, number of factors that are perfect squares = 3 (excluding 1)

For 3^2, number of factors that are perfect squares = 1 (excluding 1)

For 5^4, number of factors that are perfect squares = 2 (excluding 1)

let's use combination now :

You have total six numbers and each of them is a square in itself . so one of them or group of any combination up to all numbers are also perfecet square . In other words, out of 6 in total we, can select one of them, two of them and so forth up to all (six) of them and also "ONE" as it is not included in our calculation.

so 6C1 + 6C2 +6C3 +6C4 +6C5 +6C6 + 1 = 64 (2^6)

• I think we end up with double counting. For instance, while it is true that 2^6 contains 3 different perfect square (2^2, 2^4 and 2^6), we cannot assume those 3 as a combination of possible factors. Otherwise, we end up with counting a factor multiple times.

54,00,000 can be written as (2^6)*(3^3)*(5^5)

To form the perfect square we should consider (2^6)*(3^2)*(5^4)

since the general formula to find the number of factors is (a+1)*(b+1)*...and so on

to find the number of divisor being perfect square the formula reduces to
(a/2 +1 ) * (b/2 +1) ....so on --> just think again you will release why?

thus number of perfect square divisors are
(6/2 +1 ) * (2/2 +1) *(4/2 +1) = (3+1)*(1+1)*(2+1)
= 4*2*3
= 24

Solution:
5400000=2^6*3^3*5^5
this can be simplified for the power of perfect square
[(2^2)^3]*[(3^2)^1]*[(5^2)^2]
taking the power of all the perfect squares above
we have
3,1,2
now we know the formula to find the number of divisors is (x+1)(y+1)(z+1)......
so putting the value we have (3+1)*(1+1)*(2+1)=4*2*3=24

thanx

• The number of perfect-square factors is 24.

First, we know that 5,400,000 = 2^6 x 3^3 x 5^5. Then, take out the set of prime factors with even power. We end up with 6 factors (2^2, 2^4, 2^6), (3^2) and (5^2, 5^4).

From this point, we will calculate the combinations of possible factors out of those 6 factors.

1-type prime factor = 6 + 1 = 7
2-type prime factor = 3C1 x 1C1 + 3C1 x 2C1 + 1C1 x 2C1 = 3 + 6 + 2 = 11
3-type prime factor = 3C1 x 1C1 x 2C1 = 6
====================================================
Total number of possible perfect square factors = 7 + 11 + 6 = 24