Problem Solving

Can someone suggest an easiest way to solve this problem?

there are four consecutive prime numbers in ascending order. the product of the first three is 4199 and of the last three is 7429. what would be the largest number and what would be the smallest number of the four numbers?

Well, you know that

a*b*c = 4199

You also know that

b*c*d = 7429

Therefore, 4199*d/a = 7429

At this point, I'd backsolve, starting with answer (C). I could save time by quickly calculating ratio of d/a, which is just over 1.5. Any answer choice that wasn't close to this I could eliminate.

Same as what cbenk did, just a little more elaboration. Division is probably the easiest way I can think of solving this, afterall we are dealing with primes! Plus the question provides you with a hint that of the four prime numbers, two are common to each of the products represented.

Let the prime numbers be: A, B, C, and D. From the question we know,
A*B*C=4,199...(i)
B*C*D=7,429...(ii)

Dividing (i) by (ii),
[(A*B*C)/(B*C*D)]=[4,199/7,429]
[A/D]=[4,199/7,429]= 13/23

Since they are primes, you can safely state that the smallest number is 13 and the largest is 23.

Note:
4,199=13*17*19
7,429=13*17*19*23

Doesn't take that long to figure out the factorization, I started with 7, and found a factor when I got to 13. Once I figured this out, the rest of the math is really not neessary since 13 was the first prime factor (lowest as well) and the question tells you that the numbers are a product of consecutive primes. Naturally, feel free to double check on the test if you have time but you should also have some level of confidence that if your starting logic is correct then there really is no need to do all the math. So, once you figure out that 4,199=13*_*_, so can safely assume that the blanks are 17 and 19. Same thing with 7,429 since the first number of the product is the same as for 4,199. Hope this makes sense.

Great solutions.

This question can be done fairly quickly via backsolving and brute force (which is why I doubt that it's a real GMAT question).

On the GMAT, we'll always have 5 answer choices to help us out. First we use common sense to see if we can eliminate any choices. Even if we can't, we'd simply start with either B or D and have a 40% chance of getting it right on the 1st try and a 100% chance of getting it right on the 2nd.

For example, let's say the choices were:

A) 7, 17
B) 11, 19
C) 13, 23
D) 17, 29
E) 19, 31

Let's start with common sense: for D, the last 3 primes give us 19*23*29. Well, this is certainly more than 20^3, which is 8000, way too much (we want 7429). E is even bigger: eliminate D and E.

A, B and C are all in the realm of possibility (although you could also use common sense and a tiny bit of math to see that A is clearly going to be too small), so let's try B, the middle remaining choice:

11, 13, 17, 19

11*13*17 (a bit of brute force long multiplication never hurt anyone) = 2431. It's supposed to equal 4199, so clearly we need bigger numbers: eliminate A and B, choose C.

We also could have eliminated A and B just by focusing on the units digits:

For A, our first 3 numbers are 7, 11, 13. 7*1*3 ends in "1"; we want our product to end in "9".

For B, our first 3 numbers are 11, 13, 17. Again, 1*3*7 ends in "1"; we want our product to end in "9".

For C, our first 3 numbers are 13, 17, 19. 3*7*9 ends in "9"... bingo!

Very good solutions

I did mine a bit differently.

we can say a,b,c,d are 4 consecutive prime numbers.

a*b*c = 4199
b*c*d = 7429

by a quick look we can see that both numbers are not divisible by 2,3, or 5 since the numbers are not even, do not add up to a multiple of 3 or don't have a 5 or 0 at the end of their units digit. So the smallest number must at least be >5

b*c*d - a*b*c = b*c(d-a) = 7429 - 4199

b*c(d-a) = 3230; 3230 is divisible by 10 here. We know that in the 4 consecutive numbers there is neither a 2 or 5, which means b*c does not make a 10.

The only possibility is that d - a makes 10; which means the 4 consecutive prime numbers are at least 10 numbers apart.

13, 17, 19, 23 fit the puzzle here.

Thanks guys for the brilliant solutions. Mine was the brute force one and it took me some time so I was searching for a faster way of doing it.

Thanks!!

punitkaur wrote:there are four consecutive prime numbers in ascending order. the product of the first three is 4199 and of the last three is 7429. what would be the largest number and what would be the smallest number of the four numbers?

Given:
1) p1<p2<p3<p4 (we need to find values of p1 and p4)
2) p1 x p2 x p3 = 4199
3) p2 x p3 x p4 = 7429

Dividing 3) by 2) we get:
p4/p1 = 7429/4199

HCF of 7429 and 4199 (as determined by the division method) = 323

thus, p4/p1 = 7429/4199 = (323 x 23) / (323 x 13)
i.e. p4/p1 = 23/13

Since both 23 and 13 are primes, these must be the values we are looking for.

i.e. p4 = 23 and p1 = 13


(as an aside, 323 above is nothing but the value of(p2 x p3) which further means that p2 = 17, and p3 = 19... this does not need to be solved though for the question asked above)