Mary chose an even 4-digit number n. She wrote down all the divisors

Question

Mary chose an even 4-digit number n. She wrote down all the divisors of n in increasing order from left to right: 1, 2, ..., n/2, n. At some moment Mary wrote 323 as a divisor of n. What is the smallest possible value of the next divisor of n written to the right of 323?

(A) 324
(B) 330
(C) 340
(D) 361
(E) 646

Are You Up For the Challenge: 700 Level Questions

(A) 324
(B) 330
(C) 340
(D) 361
(E) 646

Are You Up For the Challenge: 700 Level Questions

GMATist1 · Answer

n = 4 digit even number
n =  2^{a}  * b ; where a & b are integers; Also, a>0

Now,
323 = 17 * 19

Observe: -
17*19 is part of "b" in the representation of "n" above.
 323 does not have a factor of "2" but n surely does.

Smallest possible value of next divisor = 323 * 2 = 646

Hence, E

CEO2021 · Answer

343= 17*19

So next will be either any multiple of 17 or 19 but greater than 343

19*18 =340
or consider 17*20= 340

Ans C

ThatDudeKnows · Answer

If we look at the answer choices, we can see that multiplying any of them by 323 will result in a product that is larger than four digits. That means our answer must share a factor with 323. 323 factorizes into 17 and 19. 323+17=340 would be the smallest option.

Answer choice C.

Regor60 · Answer

Call the 4 digit number, N.

We know 323*some factor K1 = N

We want the minimum number X above 323 that multiplied by its factor K2 also equals N.

So N=323K1 = X*K2

We want to minimize X and maximize K2.

Let K2 then equal K1 minus an increment L. We want to minimize L.

So 323K1 = X(K1-L), so

X= 323K1/(K1-L). 323 = 17*19 so

17*19K1/(K1-L) = X

If we want to assume L=1 for the sake of minimization, K1 can't be divisible by (K1-1) unless K1=2, which it can't be because K1 has to be at least 4 in order to create a 4 digit number.

So that leaves 17 or 19 to be divisible by K1-1. So

K1= 18 or 20. Trying both

19*17*18/17 = 19*18 = 342

19*17*20/19 = 17*20 = 340

So the minimum X, the next factor higher than 323 is 340

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Kinshook · Answer

Given: Mary chose an even 4-digit number n. She wrote down all the divisors of n in increasing order from left to right: 1, 2, ..., n/2, n. At some moment Mary wrote 323 as a divisor of n.
Asked: What is the smallest possible value of the next divisor of n written to the right of 323?

323 = 17*19
(A) 324; 324 = 2^2*3^4; Since there is no common factor between 323 & 324, n must be at least 323*324 = 104652; Since n is a 4-digit number not feasible
(B) 330; 330 = 2*3*5*11; Since there is no common factor between 323 & 330, n must be at least 323*330 = 106590; Since n is a 4-digit number not feasible
(C) 340; 340 = 2^2*5*17; n must be a multiple of 2^2*5*17*19 = 6460; If n = 6460; Feasible
(D) 361
(E) 646

No need to check futher since the smallest possible value of the next divisor of n written to the right of 323 is found.

IMO C

samarpan.g28 · Answer

Factors of 343 are 1,17,19,343.
If one divisor is 17*19 then the next can be 17*20 i.e. 340. Option (C) is correct.

KarishmaB · Answer

The first challenge here is to split 323 into its factors. We know that 18^2 =324 so we must check for all prime factors till 18. We see that 
323 = 17*19
This means that n has at least 17, 19 as factors.

Method 1: Logic

The next factor needs to have some common factor with 323 because if it had no common factors then n would have all factors of 323 and this next factor (say 324) and so n would have more than 4 digits (because 323*324 has more than 4 digits).

So the next smallest factor greater than 323 must be 17*20 =340. 
Note that 18*19 = 342 which is greater than 340.

Method 2: Use Options

If the next factor to the right of 323 were  324 = 2^2 * 3^4 
Then n would have at least 2^2*3^4 as other prime factors with these exponents.
But 323*324 has more than 4 digits so this is not possible.

If the next factor to the right of 323 were  330 = 2 * 3 * 5 * 11 
Then n would have at least 2*3*5*11 as other prime factors with these exponents.
But 323*330 has more than 4 digits so this is not possible.

If the next factor to the right of 323 were  340 = 2^2 * 5 * 17 
Then n would have at least 2^2*5 as other prime factors with these exponents.
323*20 has 4 digits so this is possible. Since it is the smallest of the options which is possible, this is the answer.

Answer (C)

Kinshook · Answer

Mary chose an even 4-digit number n. She wrote down all the divisors of n in increasing order from left to right: 1, 2, ..., n/2, n. At some moment Mary wrote 323 as a divisor of n.

What is the smallest possible value of the next divisor of n written to the right of 323?

323 = 17*19

2 is also a factor

n = 2*17*19*x where x<16 since n is 4-digit number

Next divisor of n may be either a multiple of 17 or 19

17*20 = 340
19*18 = 342

If x = 10; 17*20 = 340 may be next divisor of n.

IMO C

Mary chose an even 4-digit number n. She wrote down all the divisors

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Mary chose an even 4-digit number n. She wrote down all the divisors

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