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Three is the largest number that can be divided evenly into 27 [#permalink]
Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100

A. 30
B. 70
C. 210
D. 300
E. 700
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Three is the largest number that can be divided evenly into 27 [#permalink]
Three is the largest number that can be divided evenly into 27 and the positive integer x
So x contains 3

10 is the largest number that can be divided evenly into both 100 and x
So x contains 2 and 5

So x = 2.3.5.n

largest possible number that could be divided into x and 2100
2100= 3*2*2*5*5*7
x = 2.3.5.n

So largest no should be 300=3*5*5*2*2 as it contains 2,3 and 5

What am I doing wrong???
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Re: Three is the largest number that can be divided evenly into 27 [#permalink]
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Hi anurag16,

This question can be dealt with in a couple of different ways, but it ultimately comes down to Prime Factorization.

We're told that the largest number that divides into 27 and X is 3.

Since 27 = (3)(3)(3), that means that X's prime factorization can contain JUST ONE 3 (although it can contain other prime factors). If it contained more than one 3, then "3" would NOT be the largest number that would divide into 27 and X.

So X could be 3, 6, 12, 15, 30, etc.

Next, we're told that the largest number that divides into 100 and X is 10.

Since 100 = (2)(2)(5)(5), that means that X's prime factorization can contain JUST ONE 2 AND JUST ONE 5 (although it can contain other prime factors).

So we know that X's prime factorization consists of ONE 2, ONE 3 and ONE 5 and possibly some other primes.

We're asked for the LARGEST number that could divide into X and 2100.

2100 = (3)(7)(2)(2)(5)(5). Using what we know about X, the LARGEST number that could divide X AND 2100 would be (2)(3)(5)(7) = 210.

Final Answer:

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Three is the largest number that can be divided evenly into [#permalink]
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smartass666 wrote:
Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100

A. 30
B. 70
C. 210
D. 300
E. 700


This question is based on very simple concepts but the wording is a bit sneaky. We will come to that in a minute.
First, let's quickly analyse what we are given:
3 is the largest number that can be divided evenly into 27 ( = 3^3) and the positive integer x --> 27 has three 3s but only one 3 is common to 27 and x. It means x has exactly one 3.
10 is the largest number that can be divided evenly into both 100 (= 2^2 * 5^2) and x --> Only one 2 and one 5 is common between 100 and x so x has exactly one 2 and exactly one 5.
So x has exactly one 3, one 2 and one 5.

Which of the following is the largest possible number that could be divided into x and 2100?
We need to find the largest possible number that could be divided in x and 2100. Note that x cannot have more than one 3, 2 and 5.
2100 = 2^2 * 3 * 5^2 * 7
It has two 2s and two 5s. But x cannot have two 2s and two 5s - these are two of the limiting conditions we have on x.
What about 7? The questions doesn't tell us whether x has 7. But could it have 7? Sure. We have no limiting condition on having other factors. So the maximum common factors that 2100 and x could have are 2 * 3 * 5 * 7 = 210
So 210 is the largest possible number that could be divided into x and 2100.

Answer (C)

Check out this post: https://anaprep.com/number-properties-r ... e-factors/

Originally posted by KarishmaB on 12 Oct 2015, 23:12.
Last edited by KarishmaB on 08 Aug 2023, 04:07, edited 1 time in total.
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Three is the largest number that can be divided evenly into [#permalink]
This question is tricky because of the convoluted language. So x has one 3, one 5 and one 2 as prime factors, and also could have some other primes as no contraints have been set: x = 3*5*2*n. That is 30, 30*7, 30*11, 30*17...

2100 = 7*3*5*2*5*2. These two numbers obviously share such primes as 3*5*2. Since x cannot have more 3s, 2s, and 5s as prime factors we can borrow one 7 from 2100 to construct our max possible common factor of the 2100 and x = 30*7=210.
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Re: Three is the largest number that can be divided evenly into [#permalink]
five is the largest number that can divide evenly into 25 and the positive integer k. While 9 is the largest number that can divide evenly into both 81 and k. Which of the following is the largest possible number that could be divided into k and 270?

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Three is the largest number that can be divided evenly into [#permalink]
KarishmaB wrote:
smartass666 wrote:
Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100

A. 30
B. 70
C. 210
D. 300
E. 700


This question is based on very simple concepts but the wording is a bit sneaky. We will come to that in a minute.
First, let's quickly analyse what we are given:
3 is the largest number that can be divided evenly into 27 ( = 3^3) and the positive integer x --> 27 has three 3s but only one 3 is common to 27 and x. It means x has exactly one 3.
10 is the largest number that can be divided evenly into both 100 (= 2^2 * 5^2) and x --> Only one 2 and one 5 is common between 100 and x so x has exactly one 2 and exactly one 5.
So x has exactly one 3, one 2 and one 5.

Which of the following is the largest possible number that could be divided into x and 2100?
We need to find the largest possible number that could be divided in x and 2100. Note that x cannot have more than one 3, 2 and 5.
2100 = 2^2 * 3 * 5^2 * 7
It has two 2s and two 5s. But x cannot have two 2s and two 5s - these are two of the limiting conditions we have on x.
What about 7? The questions doesn't tell us whether x has 7. But could it have 7? Sure. We have no limiting condition on having other factors. So the maximum common factors that 2100 and x could have are 2 * 3 * 5 * 7 = 210
So 210 is the largest possible number that could be divided into x and 2100.

Answer (C)


KarishmaB can we say that because of the word COULD our answer is (C) and not (A)? Had the question been "which of the following MUST divide into" then (A) would be our answer since we are 100% sure that x will have one 3 and one 10 BUT not sure if it will have a 7,
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Three is the largest number that can be divided evenly into [#permalink]
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Hoozan wrote:
KarishmaB wrote:
smartass666 wrote:
Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100

A. 30
B. 70
C. 210
D. 300
E. 700


This question is based on very simple concepts but the wording is a bit sneaky. We will come to that in a minute.
First, let's quickly analyse what we are given:
3 is the largest number that can be divided evenly into 27 ( = 3^3) and the positive integer x --> 27 has three 3s but only one 3 is common to 27 and x. It means x has exactly one 3.
10 is the largest number that can be divided evenly into both 100 (= 2^2 * 5^2) and x --> Only one 2 and one 5 is common between 100 and x so x has exactly one 2 and exactly one 5.
So x has exactly one 3, one 2 and one 5.

Which of the following is the largest possible number that could be divided into x and 2100?
We need to find the largest possible number that could be divided in x and 2100. Note that x cannot have more than one 3, 2 and 5.
2100 = 2^2 * 3 * 5^2 * 7
It has two 2s and two 5s. But x cannot have two 2s and two 5s - these are two of the limiting conditions we have on x.
What about 7? The questions doesn't tell us whether x has 7. But could it have 7? Sure. We have no limiting condition on having other factors. So the maximum common factors that 2100 and x could have are 2 * 3 * 5 * 7 = 210
So 210 is the largest possible number that could be divided into x and 2100.

Answer (C)


KarishmaB can we say that because of the word COULD our answer is (C) and not (A)? Had the question been "which of the following MUST divide into" then (A) would be our answer since we are 100% sure that x will have one 3 and one 10 BUT not sure if it will have a 7,


Yes Hoozan
Your analysis is correct.

Check out this post on factors too: https://anaprep.com/number-properties-r ... e-factors/

and these videos on Factors and Factorisation:
https://www.youtube.com/watch?v=DxIH8rjhpKY
https://www.youtube.com/watch?v=Kd-4cH4cqHw
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