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25 Feb 2026, 20:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:26) correct
40%
(02:41)
wrong
based on 151
sessions
History
Date
Time
Result
Not Attempted Yet
What is the smallest positive integer that must be added to 1,000 so that the resultant number leaves remainders of 1, 2, and 4 when divided by 4, 5, and 7 respectively?
A. 97
B. 107
C. 117
D. 127
E. 137
A. 97
B. 107
C. 117
D. 127
E. 137
26 Feb 2026, 06:00
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Using options only
1097÷4 gives remainder 1,
1097÷5 gives remainder 2,
1097÷7 gives remainder 5. Eliminated
1107÷4 gives remainder 3,
1107÷5 gives remainder 2,
1107÷7 gives remainder 1. Eliminated
1117÷4 gives remainder 1,
1117÷5 gives remainder 2, and
1117÷7 gives remainder 4.
Ans = C
1097÷4 gives remainder 1,
1097÷5 gives remainder 2,
1097÷7 gives remainder 5. Eliminated
1107÷4 gives remainder 3,
1107÷5 gives remainder 2,
1107÷7 gives remainder 1. Eliminated
1117÷4 gives remainder 1,
1117÷5 gives remainder 2, and
1117÷7 gives remainder 4.
Ans = C
26 Feb 2026, 11:25
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Plz let me know if there's any other way to solve it even quicker.
As of now, I'd recommend backsolving this ques, starting from the smallest of the options to arrive at the correct answer in time.
If we go by back solving, it won't take much to understand that if we first 1000 to one of the options, then subtract the remainders from the formed integer one by one as per given and then divide by their respective divisors i.e. 4,5, and 7, it should form an integer.
Accordingly, we will arrive at 117 to test as one of our options.
(1000+117-1)/4 = 279
(1000+117-2)/5 = 223
(1000+117-4)/7 = 159
Since, all the above form an integer, we know it has to be correct.
Note: It's basically forming these equations:
Let x be some constant added to 1000 and Q be the quotient
Then as per our formula: (D/S)=(Q+R/S), we will get the following three equations
(1000+x) = 4Q+1
(1000+x) = 5Q+2
(1000+x) = 7Q+4
Option C
As of now, I'd recommend backsolving this ques, starting from the smallest of the options to arrive at the correct answer in time.
If we go by back solving, it won't take much to understand that if we first 1000 to one of the options, then subtract the remainders from the formed integer one by one as per given and then divide by their respective divisors i.e. 4,5, and 7, it should form an integer.
Accordingly, we will arrive at 117 to test as one of our options.
(1000+117-1)/4 = 279
(1000+117-2)/5 = 223
(1000+117-4)/7 = 159
Since, all the above form an integer, we know it has to be correct.
Note: It's basically forming these equations:
Let x be some constant added to 1000 and Q be the quotient
Then as per our formula: (D/S)=(Q+R/S), we will get the following three equations
(1000+x) = 4Q+1
(1000+x) = 5Q+2
(1000+x) = 7Q+4
Option C
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