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# Numbers and Divisiblity

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Intern
Joined: 31 Jul 2017
Posts: 1

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19 Aug 2017, 12:41
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Let N be the greatest number that will divide 1305,4665 and 6905, leaving the same reminder in each case.Then sum of digits in N is :

A. 4
B. 5
C. 6
D. 7
E. 8
BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 2256
Location: India
GPA: 3.12

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19 Aug 2017, 22:35
1
KUDOS
Hi sarth123,

Welcome to GMATClub!

In order to solve the question, you must be aware of the following rule
If the remainder is same in each case and remainder is not given,
HCF of the differences of the numbers is the required greatest number

The three numbers $$1305, 4665$$ and $$6905$$ will have differences $$(4665 - 1305 = 3360),(6905 - 4665 = 2240),$$ and $$(6905 - 1305 = 5600)$$

In order to find the greatest number which divides each of these differences,
we prime factorize the differences
$$3360 = 2^5 * 3 * 5 * 7$$
$$2240 = 2^6 * 5 * 7$$
$$5600 = 2^5 * 5^2 * 7$$

The HCF of these numbers is N - 1120(2^5 * 5 * 7) and is the greatest number leaving the same remainder.

Since we have been asked to find the sum of digits in N, it must be 1+1+2+0 = 4(Option A)
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Math Expert
Joined: 02 Sep 2009
Posts: 44290

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20 Aug 2017, 03:15
sarth123 wrote:
Let N be the greatest number that will divide 1305,4665 and 6905, leaving the same reminder in each case.Then sum of digits in N is :

A. 4
B. 5
C. 6
D. 7
E. 8

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Re: Numbers and Divisiblity   [#permalink] 20 Aug 2017, 03:15
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