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12 Days of Christmas GMAT Competition 2025 - Day 9: What is the

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750rest

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25 Dec 2025, 11:39

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By option Ans - A as rest opytions can't devide two different odd positive integers less than 213.

Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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AditiDeokar

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25 Dec 2025, 11:56

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The anser is A. 69. GCF of two odd numbers less than 213 would mean out of the options which number into 3 would be below 213 and also be a gcf, so 69 x 3 = 207, gcf of 207 and 69 is 69, any other tested number would be beyond 213.

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25 Dec 2025, 12:33

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Options C D or E will produce even multiples when multiplied by 2 and will be greater than 213 if multiplied by 3. So these numbers can't be the odd HCF of the integers mentioned in the question that must be less than 213.
Option B if multiplied by 3 becomes 213, so this can't the be the highest common factor as well.
So the remaining answer is the correct one.

A

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25 Dec 2025, 13:32

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Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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We need gcd of two different odd positive integers, less than 213.

So it should be in ratio of x : 3x

Let's try 71:-

71 : 213 ( Not possible as it should be less than 213)

Our answer is 69 - A

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25 Dec 2025, 13:48

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Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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I plugged in the options. in the given options.. we can eliminate 207 and 209..as it is too large.

105 x 2 is even, 105 x 3 is out of scope

71 x 2 is even, 713 is not in scope..as we need number less than 71.

69 x 3 is 208. ..208 can be divided. by 69,3.

Hence the answer is 69

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25 Dec 2025, 14:59

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To maximize the GCD off odd numbers less than 213 we need the smallest co-prime and multipliers which in this case we can say x=1 and y=3 while a is the gcd
So we can have the numbers as a and 3a
Given that 3a is the bigger one and has to be less than 213, we can simplify to
3a<213
a<71
The next lower odd number from 71 is 69 meaning the other number is 2007
So the gcd of both is 69
Ans A

Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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obedear

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We are essentially being asked; what is the greatest positive factor which can divide two different positive odd integers. Since we are being asked to maximize it, we need to consider high numbers. First of all, every number is a factor of itself, which means one of those divisors is already taken care of. The smaller the prime factor we divide the number by, the larger the other factor will be, so we should focus on dividing the high odd number by 3 (not 2, since that would mean it is an even integer).

Testing answer choices: 71*3 = 213 which does not fit the question stem. Moving down by odd intervals. 211 is not divisible by 3 and neither is 209 so we can move to 207 which divided by 3 = 69.
Can't be anything greater than that, so 69 is the answer.

69 is the answer.

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Given that the two positive integers will be less than 213 and will be odd. So , the highest common divisor will be something that on multiply with 3 gives a number in the ranges of 200. I know the question will be why. It's because the numbers cannot be even first. So multiplying factor two cannot be possible. Multiplying should be small and common divisor should be greatest to get greatest final result.

So , let's start with 209. 209 is divisible by 11/19. Rule this out.

Let's take 207 , 207 is divisible by 69 and 3. Bingo so , 69 will be the answer. Let's go deeper and will check if we are something. But now as we go lesser , divisor value will be getting smaller.

Let's take 205 , it's divisible by 41 and 5. so 41<69. Even not divisible by 3. Let's take 201. 201 is divisible by 67 and 3. 67<69.

So Final answer is A.

Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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Let the greatest common divisor between the two integers be g
Since both the numbers are odd, gcd is also odd.

To maximise g, we need find smallest multipliers of g like 1,3,5,7 etc
For two distinct numbers, 1 cannot be taken.

Taking the multiplier as 3,
3g <213
g<71 and maximising g with constraint as odd gives
g = 69

The numbers are 69 and 207

So the greatest common divisor of 69 and 207 is 69

Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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We are to find two numbers whose HCF would be as high as possible and the numbers themselves should be less than 213. But the biggest constraint is that both the numbers are to be odd.

So it cannot be 105 and 210 as one of the number is even.

Nearest odd number to 213 divisible by 3 is 207 which can be represented as 69*3 so 69 and 207 can be the two numbers having a common divisor as 69 itself. 71 and 213 would not be applicable as numbers are to be less than 213. No other option would satisfy our constraints either.

Therefore, Option A imo

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the numbers should be positive,odd and less than 213
so by option the common divisor can be 69
because the two nos can be 69 and 207
which are odd and less than 213 and positive hence answer is 69

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Let the GCD be x, the 2 integers must be multiples of x
Since neither of these numbers is even, the numbers can be odd multiples (ie, x and 3x)

But 3x < 213
x < 71

Thus, in the options only 69 satisfies this.

Answer: A

Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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25 Dec 2025, 23:57

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First, note that both are odd and distinct.
For 209 to be the GCD of two distinct odd integers, they have to be odd multiples of 209. Say 209(minimum) and 627(209 * 3, next smallest odd multiple), but both have to be less than 213.
So we can eliminate 209, 207, 105, 71, because the next minimum odd multiple for them is not less than 213.
For 69, the next odd multiple is 69*3 = 207, less than 213.

Option A.

Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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Two number is less than 213, both are odd => the number is aroudn 211, 209 => it may be prime numbers, but we first leave them there.
My approach is to play around with the options, to find the common division for the options
A. 69 - 69 may be a commond division for 69, and 69*3=207
B. 71 - 71*3=213 => eliminate
C. 105 - too large if times with 3 => eliminate
D,E. eliminate because it might be prime number and cannot be common division for number less than 213
A

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Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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In this case, we need to find if at least two odd multiples of the options provided are less than 213

Option A. 69 * 1 = 69, 69 * 3 = 207 (Satisfies)
Option B. 71 * 1 = 71, 71 * 3 = 213. Since the positive integers must be less than 213. This does not satisfy the condition
Similarly, the options C, D and E will not satisfy, as if you multiply each of them by 3, the result will always be greater than 213

Hence, the answer is A

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Scan the options: I'm gonna check multiply each options with 1,3 since the integers must be odd.
A. 69. 69*3=207. So GCM(207,69) = 69. Temporarily, this satisfies each integers < 213 and odd.
B. 71. 71*3 = 213. This is not less than 213 so eliminate.
C. 105. 105*3 = 315. Eliminate.
Obviously eliminate D and E since GCM cannot be close to 213.

Answer: A

Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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GCD of 2 different +ve integers less than 213 (both odd)
a<b<213
K1=1
K2=3
b=3g<213
g<213/3
g<71
so less than 71 and also odd=69

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26 Dec 2025, 03:06

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If 2 numbers have a GCD, d, they can be written as d and kd where both d and k should be odd (both numbers are odd in the question)
The largest possible odd as per the question also has to be less than 213.

i.e. => 1*d< 213 and k*d < 213

To maximize => k =3 (smallest, odd number) => d< 213/3 => d<71

Based on the answer choices the best answer seems to be 69 => Option A

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26 Dec 2025, 04:16

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Answer C

of the choices

a 69, 138
b 71, 142
c 105. 210
d 207, 414
e 209, 418

the largest and still less that 213 is C 105, 210

Bunuel

What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

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What is the greatest possible common divisor of two different positive integers, each less than 213, if neither of them is even?

A. 69
B. 71
C. 105
D. 207
E. 209

to find greatest possibile HCF

Best way to solve is to try each answer....

A. 69 ---- given number is already high, to get the second number, we try to check by multiplying by 2, or 3

Possible Numbers = 138, 207.

Possible pair of Numbers - {69, 138} , {69,207} , {138, 207}

Correct pair without even number {69,207} (Answer)

B. 71 ----------------to find possible answer on multiply by 2.

Possible Numbers = 142, 213.

No possible pair. (either even or equal to 213)

C. 105 ------------to find possible answer on multiply by 2.

Possible Numbers = 210, 315

No possible pair. (either even or grater than 213)

D. 207 --------
Possible pair = 414

All further numbers will be greater than 213.