If a, b, c and d are integers such that 1

Question

If a, b, c and d are integers such that 1 < a ≤ b ≤ c ≤ d and abcd = 660, then how many possible combinations exist for value of a, b, c, d ?

A. 0
B. 1
C. 2
D. 5
E. 7

PyjamaScientist · Answer

Option (D)

1<a<= b <= c <= d
abcd = 660 = 2^2 * 3 * 5 * 11
So, a can be either 2 or 3.
When a = 2,
2, 2, 3, 55
2,2, 11, 15
2, 3, 10, 11
2, 5, 6, 11
When a = 3,
3, 4, 5, 11

Total 5 combinations.

RanonBanerjee · Answer

IMO C.

prime factorization of 660 =2×2×5×3×11

We need to make 4 numbers out of these 5, and naturally there would be an order where 1<a<=b<=c<=d

So total number of combinations possible is 5C4=5.

Posted from my mobile device

IanStewart · Answer

The answer is 7. We have the prime factorization (2)(2)(3)(5)(11), where we have five primes (two of them identical) from which we need to make four numbers, using all of these primes. So really we're going to multiply two of the primes together, and leave the other three primes alone. So the question is how many distinct products we can make if we'll multiply two of these primes together. If the primes were all different, the answer would just be 5C2 = 10. But we have the repeated '2', and we don't want to count a product like 2*3 twice, once for each '2'. We only want to count that product once. Since we don't want to double-count the products 2*3, 2*5, 2*11, which is three products in total, the answer is 10 - 3 = 7. Or we could just list all seven possibilities, but it's important to be systematic:

2, 2, 3, 55
2, 2, 5, 33
2, 2, 11, 15
2, 3, 5, 22
2, 3, 10, 11
2, 5, 6, 11
3, 4, 5, 11

Bunuel - is this your question? The word "combinations" has a reserved meaning in combinatorics, and this is not technically speaking a combinations questions as it's worded.

2, 2, 3, 55
2, 2, 5, 33
2, 2, 11, 15
2, 3, 5, 22
2, 3, 10, 11
2, 5, 6, 11
3, 4, 5, 11

Bunuel - is this your question? The word "combinations" has a reserved meaning in combinatorics, and this is not technically speaking a combinations questions as it's worded.

sanjitscorps18 · Answer

1 < a ≤ b ≤ c ≤ d and abcd = 660
Now 660 = 2^2 x 3 x 5 x 11

Since we have two 2s we can play around that

Possible combinations with 2 multiplied with each prime once
 4  x 3 x 5 x 11
2 x  6  x 5 x 11
2 x 3 x  10  x 11
2 x 3 x 5 x  22

Total 4 combinations

Now we keep both the 2s as individual factors and manipulate the other numbers
2 x 2 x  15  x 11
2 x 2 x 3 x  55 
2 x 2 x 5 x  33

We get 3 more combinations

Total 7 combinations

Option E

Jatro · Answer

As it will select 4 cases out of 5, and here we will need another 2 that is repeating as well

1) Total cases : 4

2,3,5,11 - 2 can multiply each of the prime numbers

2) One pair of prime numbers multiply each other to fit in 4 numbers, as we have already used 2 in the above case we will not include here

3 cases exist here

2 x 2 x 3 x 55

2 x 2 x 11 x 15

2 x 2 x 5 x 33

Kinshook · Answer

Asked: If a, b, c and d are integers such that 1 < a ≤ b ≤ c ≤ d and abcd = 660, then how many possible combinations exist for value of a, b, c, d ?

abcd = 11*2^2*3*5

Since a, b, c & d are not 1 and they are all integers.

Possible combinations = {(2,2,3,55),(2,2,15,11),(2,2,5,33),(2,3,5,22},(3,4,5,11},(2,6,5,11},(2,3,10,11}} : 7 combinations are possible.

IMO E

If a, b, c and d are integers such that 1 < a ≤ b ≤ c ≤ d and abcd = 6

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