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19 Nov 2020, 02:30
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B
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Difficulty:
75%
(hard)
Question Stats:
56% (02:40) correct
44%
(02:29)
wrong
based on 178
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How many three-digit positive integers are there such that when divided by 7 or 8 give a remainder of 4?
A. 15
B. 16
C. 17
D. 18
E. 19
A. 15
B. 16
C. 17
D. 18
E. 19
19 Nov 2020, 05:57
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Bunuel
LCM of 7 and 8 = 56
Smallest 3 digit number (> 100) divisible by 56 = 112
To get a remainder of 4, the number we need is 112 + 4 = 116
Largest 3 digit number divisible by 56 is 952. To get a remainder of 4, the number is 952 + 4 = 956
The number of multiples of 56 = \(\frac{Largest \space Number - Smallest \space Number}{Divisor} + 1 = \frac{956 - 116}{56} + 1 = \frac{840}{56} + 1 = 15 + 1 = 16\)
Option B
Arun Kumar
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19 Nov 2020, 06:14
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LCM of 7 & 8 is 56
We are looking for a 3 digit number, which, when divided by 56 give leave a remainder of 4. SO it will be of a form 56x + 4, in which x is an integer.
The smallest such number is 112+4=116 = (56*2)+4
The largest such number is 952+4=956 = (56*17)+4
So the total such numbers will be 17-2+1 = 16
Answer Option B
We are looking for a 3 digit number, which, when divided by 56 give leave a remainder of 4. SO it will be of a form 56x + 4, in which x is an integer.
The smallest such number is 112+4=116 = (56*2)+4
The largest such number is 952+4=956 = (56*17)+4
So the total such numbers will be 17-2+1 = 16
Answer Option B
