A positive integer ‘A’ is a multiple of 180 and it has 40 factors. If

Question

A positive integer ‘A’ is a multiple of 180 and it has 40 factors. If ‘A’ is less than 3000, then the value of A/40 is

(A) 54
(B) 60
(C) 240
(D) 270
(E) Cannot be determined

nick1816 · Answer

A=2^2*3^2*5*k

Only possible factorization of 40 is 5*4*2

2 possible values of A

1.   2^4*3^3*5=80*27<30*100=3000

2.   2^3*3^4*5=40*81>40*80>3000  ( reject it )

\frac{A}{40}= \frac{2^4*3^3*5}{2^3*5}= 2*3^3=54

MathRevolution · Answer

Let  A = 180 = 2^2 * 3^2 * 5 = (2 + 1)(2 + 1)(1 + 1) = 3 * 3 * 2 = 18    factors .

For 40 factors , we should have 4 * 5 * 2 or 5 * 4 * 2.

=>  A = 2^3 * 3^4 * 5 = (3 + 1)(4 + 1)(1 + 1) = 180 * 2 * 9 = 4 * 5 * 2 = 40   factors =  3,240 > 3000 - NOT POSSIBLE

OR

=>  A = 2^4 * 3^3 * 5 = (4 + 1)(3 + 1)(1 + 1) = 5 * 4 * 2 = 40   factors =  2160 < 3000 - POSSIBLE

=>  \frac{A}{40} = \frac{2160 }{ 40} = 54

Answer A

Fdambro294 · Answer

Step (1)
We need to find a multiple of 180, so first prime factorize 180:

180 = (2)^2 * (3)^2 * (5)

Step (2)
We are told that the multiple or 180 has 40 factors.

In order to be a multiple of 180, the number must have the same prime factors in at least the same amount as 180 (the exponents of the prime Factorization must be equal to or greater than the exponents of 180’s Prime Factorization)

Rule: to find the number of factors of a number, we take the exponent of each prime factor, add + 1, then multiply the powers

Since 180 has three distinct prime factors, we know that the multiple’s Prime Factorization must have at least:

Two (2)’s

Two (3)’s

One (5)

Since we are multiplying the integer exponents, we can break the product of 40 down into 3 factors (1 exponent for each unique prime factor) to see which exponents we will need ——-> 40 = 5 * 4 * 2

Thus, if we made the unique prime factors of 180 (2 and 3 and 5) have the following exponents:

(5-1) = 4
(4-1) = 3
(2-1) = 1

We can assure that the value would be a multiple of 180 (since the exponents of each prime factor exceeds those of 180’s prime factorization) and we would have 40 factors.

Finally, to meet the constraint that the Multiple of 180 must be less than 3,000 ——> we should assign the higher exponents to the lower value Primes

Make the Multiple of 180 equal to——-> (2)^4 * (3)^3 * (5)^1

(Step 3) check if the number meets the constraint

Number of unique positive factors or this number value = (4 + 1) (3 + 1) (1 + 1) = 5 * 4 * 2 = 40

AND

the value of the number = (2)^4 * (3)^3 * (5)^1 = 16 * 27 * 5 = 2,160

Which is less than < 3,000

(Step 4) divide this multiple of 180 by 40

(2)^4 * (3)^3 * (5)
_______________
(2)^3 * (5)

After canceling the exponents we have:

(2) * (3)^3 = 2 * 27 =

54

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SatvikVedala · Answer

since 180k<3000 k<16.6 also (180*k)/40 ==> (9/2)*k --- k has to be even for A/40 to be an integer as per options with k = 16 ----> A/40 = 72 k = 14 ----> A/40 = 63 k = 12 ----> A/40 = 54 Since as per options there is no value below 54, A is the choice

Tinytrotter · Answer

I had a completely separate approach. I just looked at the second part of the question first which asks A/40 = ?

We know that A<3,000 so instantly Option C & Option D are out.

Let us proceed with Option B, A/40 = 60 which means A=2400. But this does not work because A is a multiple of 180 and 2400/180 does not yield an integer value.

Left is only 54. If A/40 = 54 then A = 2160. This divided by 180 gives 12. We know that 180 has 18 factors (2^2 * 3^2 * 5) and now 12 can be written as 2^2*3.

So, 2160 can be expressed as 2^4*3^3*5 which gives us (5)*(4)*(2) = 40 factors.

I found my method to be the fastest, does anyone have any other faster methods?

saynchalk · Answer

Impossible to get even started for me, please help.

A positive integer ‘A’ is a multiple of 180 and it has 40 factors. If

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A positive integer ‘A’ is a multiple of 180 and it has 40 factors. If

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