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24 Feb 2020, 06:19
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
88% (01:40) correct
12%
(01:40)
wrong
based on 125
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What is the lowest positive integer which gives a remainder of 5 when divided by any of the numbers 6, 7, and 8 ?
A. 153
B. 158
C. 163
D. 168
E. 173
A. 153
B. 158
C. 163
D. 168
E. 173
24 Feb 2020, 07:36
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Bunuel
Method 1:
lowest positive integer which gives a remainder of 5 when divided by any of the numbers 6, 7, and 8 = LCM (6, 7, 8) + 5
i.e. Required Number = 168 + 5 = 173
Method 2:
Check Options
A. 153 when divided by 6 leaves remainder 3 so OUT
B. 158 when divided by 6 leaves remainder 2 so OUT
C. 163 when divided by 8 leaves remainder 3 so OUT
D. 168 when divided by 8 leaves remainder 0 so OUT
E. 173 SATISFIES REQUIREMENTS
Answer: Option E
31 Aug 2022, 01:45
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We need to find what is the lowest positive integer which gives a remainder of 5 when divided by any of the numbers 6, 7, and 8
Theory: Dividend = Divisor*Quotient + Remainder
Let's solve the problem using two methods
Method 1: Substitution
Let's take each option and check if it gives 5 as remainder when divided by 6, 7 and 8.
Before we check the option choices we can use some logic to reduce the number of options to check.
Now, we know that any multiple of 6 and 8 will always be an even number.
And the number which we are looking for is (a multiple of 6 or 8) + 5
=> The number will be Even (because of multiple of even number) + Odd (because of 5) = Odd
So, we can eliminate all Even answer choices.
=> B and D are OUT. Now lets check A, C, E
A. 153
153 when divided by 6 gives 3 remainder => OUT
C. 163
153 when divided by 6 gives 1 remainder => OUT
E. 173 -> Will be the answer as all other are out
173 when divided by 6 gives 5 as remainder
173 when divided by 7 gives 5 as remainder
173 when divided by 8 gives 5 as remainder
So, Answer will be E
Method 2: Algebra
Since the number gives 5 remainder when divided by 6, 7 and 8 that means that the number can be written as
6x + 5, 7y + 5, 8z + 5, where x, y , z are non-negative integers
=> The number is 5 more than the multiples of 6, 7, 8
=> The number is 5 more than the LCM(6,7,8)
( Watch this video to learn How to Calculate LCM of 2 or more numbers)
=> The number is 5 more than 168 = 168+5 = 173
So, Answer will be E
Hope it helps!
Playlist on Solved Problems on Remainders here
Watch the following video to MASTER Remainders
Theory: Dividend = Divisor*Quotient + Remainder
Let's solve the problem using two methods
Method 1: Substitution
Let's take each option and check if it gives 5 as remainder when divided by 6, 7 and 8.
Before we check the option choices we can use some logic to reduce the number of options to check.
Now, we know that any multiple of 6 and 8 will always be an even number.
And the number which we are looking for is (a multiple of 6 or 8) + 5
=> The number will be Even (because of multiple of even number) + Odd (because of 5) = Odd
So, we can eliminate all Even answer choices.
=> B and D are OUT. Now lets check A, C, E
A. 153
153 when divided by 6 gives 3 remainder => OUT
C. 163
153 when divided by 6 gives 1 remainder => OUT
E. 173 -> Will be the answer as all other are out
173 when divided by 6 gives 5 as remainder
173 when divided by 7 gives 5 as remainder
173 when divided by 8 gives 5 as remainder
So, Answer will be E
Method 2: Algebra
Since the number gives 5 remainder when divided by 6, 7 and 8 that means that the number can be written as
6x + 5, 7y + 5, 8z + 5, where x, y , z are non-negative integers
=> The number is 5 more than the multiples of 6, 7, 8
=> The number is 5 more than the LCM(6,7,8)
( Watch this video to learn How to Calculate LCM of 2 or more numbers)
=> The number is 5 more than 168 = 168+5 = 173
So, Answer will be E
Hope it helps!
Playlist on Solved Problems on Remainders here
Watch the following video to MASTER Remainders
