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

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Bunuel

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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

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Bunuel

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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

GMAT Club Official Explanation:

We want to find the greatest possible common divisor, say d, of two different odd positive integers, both less than 213.

To maximize d, we make the smaller number d itself and the larger number 3d (since 2d would be even, we skip it). This gives us the closest pair, allowing us to maximize d. Thus, we need such 3d that it's less than 213 and odd.

The greatest odd multiple of 3 less than 213 is 207, so 3d = 207, and d = 207/3 = 69.

Answer 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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Using Options

A. 69

First Number = 69 * 1
Second Number = 69 * 3 = 207

Hence this is possible.

B. 71

First Number = 71 * 1
Second Number = 71 * 3 = 213

The question states that the numbers are less than 213, hence any number above 71 will not be possible.

We can therefore reject B, C, D, and E.

Option A

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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

2 different odd +ve integers such that its multiple is < 213
use answer options
69*1 ; 69
69*3 ; 207
check with 71
71*1 ; 71
71*3 ; 213
not within range of question

OPTION A ; 69 is correct

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Let the numbers less than 213 be x,y
GCD(x,y) = ?

Need to find 2 odd multiples of what number is less than 213
69 is the only one that fits 69,207 as 138 (69*2) is even

Just checking for 71 as it is also a small number
For 71, we get 71,142,213 but 142 and 213 is not allowed. With any other number less than 213, it will be co-prime with GCD=1

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

Let the integers be x & y and let the common divisor be d

Since both of them are odd, lowest value of divisor multiples are 1 & 3.

x = d
y = 3d

y = 3d <213
d < 213/3 = 71

Since both of them are odd

d = 69; x = 69; y=3*69 = 207

Greatest possible common divisor of 2 different positive integers = 69

IMO A

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Greatest Possible HCF of a,b provided, none of them are even (2 cannot be the common factor)

Next possible smallest prime number = 3 (after 2) (Why? Because essentially we are dividing the HCF, hence we need a smaller divisor and 2 cannot be a divisor)

Now a = 71*3 = 213
b = 71

HCF = 71, hence B NO! Trapped (I am hence, but now I see where I faultered)

a,b<213!!! hence answer is A, 69.
a = 69*3
b = 69

Now why smallest prime number, because you are essentially dividing it by HCF so for any other number

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As both integers are odd, their GCD also must be odd. If the GCD is d, then both numbers are multiples of d.
Let numbers be d and 2d. -> but as 2d is even -> let's take numbers to be d and 3d.
Now as 3d<213 -> d<71

A - 3*69 = 207 < 213 -> so correct
B - 3*71 = 213 (not less than 213)
C - 3*105 = 315 (too large)
D - 207 > 213 so not correct
E - 209 > 213 too large

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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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For something to be the greatest common divisor, it must evenly divide into two numbers. If one of the answer choices is the GCD of a number less than 213, then it must evenly divide said number. This question essentially asks whether any of these have a multiple that's less than 213 and odd, because then the GCD divides the answer choice and the multiple evenly.

Since we can't multiply by 2, we need to find a number here that will multiply by 3 and be less than 213. We can immediately eliminate C, D and E because they will be far too big. $71*3=213$, which is not less than 213, so that can be eliminated. $69*3=207 < 213$, which is both odd and less than 213. 69 evenly divides 213 and 69. Thus, A is our answer.

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two positive integer. neither them even and both less than 213.

for 69 , its 69 and 207. both conditions met.
for 71, its 71 and 213. but we need less than 213. reject
for 1045, its 105 and 315. reject
for 207, its 207 and 621. reject
for 209 , its 209 and 627. reject

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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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 divisor be g, to maximise g
then first number will be g and second numebr wil be 3g ( since numebr can't be muleiple of 2)
3g <213
g<71
g cannot be 70 as it is even
so g = 69

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The two integers (x and y ) must be different .
Both Integers must be odd .
Both Integers must be less than 213(x, y<213)

If g is the greatest common divisor of x and y , then x and y must be multiple of g . Let x = g.a and y = g.b , a and b are different odd integers .
Smallest positive odd integers are a = 1 and b = 3
gx3 < 213 , means g<71 , Only option available here is 69 ✅

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Both distinct positive numbers are odd and both numbers must be less than 213.
If both numbers have common divisor d, then they are both multiples of d. Smallest distinct odd multipliers are 1 & 3. We need to find largest odd number d such that,
1*d<213 and 3*d<213 ; d<213/3 ; d<71
Largest possible value is 70 but we need odd value, so its 69.
Numbers are 69 and 69*3=207 and their gcd is 69

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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Need to just make the equation

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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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Let the two numbers be N1 and N2.

The greatest common divisor be denoted as h.

N1 = h*a

N2= h *b

Both a or b is NOT EQUAL to even.

Therefore, both are odd numbers.

a and b should not contain any number as common between them, which means both are co prime.

If h = 69, then a=1 and b=3 is a possible case.

69*3 = 207 < 213. Valid case.

Because, 71 multiplied with the first two odd numbers will yield 71 and 213 (which is not less than 213).

There fore, the highest common divisor is 69.

Option A .

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to have the largest gcd we have to find 2 odd integers a and b < 213 that are both multiples of the gcd
these integers could be 1*gcd, 3*gcd, 5*gcd, ... and because they cannot be even numbers, the gcd must be an odd number.
the smallest options are 1*gcd and 3*gcd

considering the restriction of a,b < 213, check which of the answers could be the option:
69*3 = 207 ->ok
71*3 = 213 -> too big
since 71*3 is already > 213, there is no point in checking the other options

the two integers must be 69, 207
Answer A. 69

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need to find the greatest possible common divisor of two different positive Integers both of which are odd and less than 213??
207 and 209 can be directly eliminated: as if one of these is 1st integer and common divisor there would be no number less than 213 that can be a multiple of it.
105 is 1st integer and common divisor then certainly the other one is 210 which is less than 213 but is even ----so out
if 71 is the 1st no. and the common divisor then only odd no. which is multiple of it is 213---so out
so 69 is the correct choice and the other no. is 207. 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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Both numbers are odd since it says neither are even,
Lets consider different options
A: 69, so the numbers should be some multiple of 69 and less than 213 and odd
lets assume its 69m and 69n
if m=1, then we have 69
if n=2, its 138 = even, so not allowed
if n=3, its 207, which is odd and less than 213.
So 69 works

for other options, we see that its odd multiples exceed 213, and the only possible value is the number itself, so we dont have another number below 213

Thus Option A is the answer

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let the numbers be a & b
a=GCD*K1
b=GCD*K2

if K1=1, K2 has to be 3.
so b<213/3<71

so b is less than 71.
here b is 69
a = 69*3 = 207
b=69

GCD=69
Ans A

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Generally,
$1 <= HCF <= min(a, b)$

so to get maximum we need a and b such b is a multiple of a.

But we have another constraint that an and b are not even. So possibilities for b are

211, 209, 207, 205 etc

it can't be 211 because it is not divisible by 3. It might be divisible by some high order number (assuming) but then any number greater than 3 will reduce a to such a value that it is less than some b/3

if we take 207 as b, 69 is a

So max HCF is 69

ans: option A