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13 Jul 2026, 09:40
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The answer is 11. Option D.
Here, the numbers are divisible by 4 and 11.
The remainder must be less than 4, i.e. 0,1,2,3.
Case 1: Remainder 0 (lcm of 4,11=44)
nos. possible- 44 and 88
Case 2: Remainder 1- should be more than a multiple of 44.
Numbers possible- 1, 45,89
Case 3: 2 more than a multiple of 44.
Numbers possible: 2, 46, 90
Case 4: 3 more than a multiple of 44.
Numbers possible: 3, 47, 91
Total= 2+3+3+3= 11 integers.
Here, the numbers are divisible by 4 and 11.
The remainder must be less than 4, i.e. 0,1,2,3.
Case 1: Remainder 0 (lcm of 4,11=44)
nos. possible- 44 and 88
Case 2: Remainder 1- should be more than a multiple of 44.
Numbers possible- 1, 45,89
Case 3: 2 more than a multiple of 44.
Numbers possible: 2, 46, 90
Case 4: 3 more than a multiple of 44.
Numbers possible: 3, 47, 91
Total= 2+3+3+3= 11 integers.
13 Jul 2026, 09:48
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When a number is divided by 4, remainder can only be 0,1,2,3.
When a number is divided by 11, remainder are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Common remainders = x= 0,1,2,3
LCM(4,11)= 44
n= 44k+x
1≤n≤100
When k=0, n=x (1,2,3 are possible values)
When k=1, n=44+x (44,45,46,47 are possible values)
When k=2, n=88+x (88,89,90,91 are possible values)
k can't be 3 or more because then n value will exceed 100.
Positive integers with common remainders are (1, 2, 3, 44, 45 ,46, 47, 88, 89, 90, 91) = 11 values
D
When a number is divided by 11, remainder are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Common remainders = x= 0,1,2,3
LCM(4,11)= 44
n= 44k+x
1≤n≤100
When k=0, n=x (1,2,3 are possible values)
When k=1, n=44+x (44,45,46,47 are possible values)
When k=2, n=88+x (88,89,90,91 are possible values)
k can't be 3 or more because then n value will exceed 100.
Positive integers with common remainders are (1, 2, 3, 44, 45 ,46, 47, 88, 89, 90, 91) = 11 values
D
13 Jul 2026, 09:51
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have many integers less than 100 have the same remainder of 4 and 11
Remainders can only be 0,1,2,3
For r=0, n can be 44, 88 ( 2 numbers)
For r=1, n can be 1, 45, 89 (3 numbers)
For r=2, n can be 2, 46, 90 (3 numbers)
For r=3, n can be 3, 47, 91 (3 numbers)
2+3+3+3 = 11
Answer D 11
Remainders can only be 0,1,2,3
For r=0, n can be 44, 88 ( 2 numbers)
For r=1, n can be 1, 45, 89 (3 numbers)
For r=2, n can be 2, 46, 90 (3 numbers)
For r=3, n can be 3, 47, 91 (3 numbers)
2+3+3+3 = 11
Answer D 11
