The HCF of two numbers is 21 and their LCM is 4641. If one of the numb

Question

The HCF of two numbers is 21 and their LCM is 4641. If one of the numbers is between 200 and 300, find the other number.

A. 221
B. 273
C. 289
D. 357
E. 441

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Showmeyaa · Answer

We know that HCF * LCM = product of two numbers
So, 21 * 4641 = number between 200 & 300 * other number
3* 7 * 3 * 7 * 221 = number between 200 & 300 * other number
 3^2 * 7^2 * 221  = 221 * other number
Other number =  3^2 * 7^2 = 441 
IMO, (E)!

ndour0291 · Answer

There is an easier approach
HCF is 21=3·7
So, each number has as factor 3 and 7
Of the a,b,c,d,e numbers
Divisible by 3 are b, d, e (sum of digits divisible by 3)
Divisible by 7 are d and e
So=> only d and e is divided by 3 and 7
Correct is either d or e

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CrackverbalGMAT · Answer

The HCF of two numbers is 21 and their LCM is 4641. If one of the numbers is between 200 and 300, find the other number?

Product of 2 numbers = HCF * LCM

N1 * N2 = 21 * 4641

Since its given that one of the number is between 200 and 300, lets try to prime factorize 21 and 4641 .

4641 = 21 * 221 = 21 * 13 *17

since one of the number is between 200 and 300 , we can write N1* N2 = 21*21 * 13 *17 = (21*17 ) *(21*13) = 357 * 273

273 is between 200 and 300 .
HCF of 357 and 273 is 21.

The correct answer is option D .357

Where as Option E is the trap answer because HCF of 441 and 221 is not 21 .

Thanks,
Clifin J Francis,
GMAT SME

HCF of 441 and 221 is

armaankumar · Answer

@

Prime Factorization of 4641 = 3*7*13*17

The question mentions the HCF is 21, so (3*7) pair will be included in both the numbers.
Two Possible Numbers: (3*7*13) and (3*7*17)
(3*7*13)= 273 
(3*7*17)= 357

Since 273 lies between 200 and 300, the correct option is D.

Aks111 · Answer

Let A and B be the two numbers.
HCF is 21 means that both A and B are multiples of 21 and there is no other common prime factor between them.

LCM is 4641 = 21* 13 * 17 : means either of below four possibilities:
1. A contains both 13 and 17 as multiples i.e. A = 13*17 *21 = 4641 and B = 21
2. B contains both 13 and 17 as multiples i.e. A = 21 and B = 13*17*21 =4641
3. A contains 13 as a multiple and B contains 17 as a multiple i.e. A = 21*13 = 273 and B = 21*17 = 357
4. A contains 17 as a multiple and B contains 13 as a multiple i.e. A = 21*17 = 357 and B = 21*13 = 273

Given one of the numbers is between 200 and 300. We see only possibility for one of the numbers to be between 200 and 300 is 273.
Hence, the other number is 357.
 Answer: D

srik410 · Answer

Because HCF is 21, we can write a=21k1 and b=21k2
 
so ab=21x4641 => k1xk2=4641/21=221=17x13
also given one number is between 200 and 300, so either k1 or k2 is between (200/21, 300/21) or

we have two primes 17 or 13. 13 falls in the above category so left = 17. hence answer= 17*21=357

The HCF of two numbers is 21 and their LCM is 4641. If one of the numb

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The HCF of two numbers is 21 and their LCM is 4641. If one of the numb

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