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13 Jul 2026, 22:21
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Let the common remainder be r.
Since the remainder when dividing by 4 must be less than 4,
r = 0, 1, 2, or 3.
A number with remainder r when divided by both 4 and 11 satisfies:
n ≡ r (mod 4)
n ≡ r (mod 11)
This means:
n − r is divisible by both 4 and 11.
Since LCM(4,11) = 44,
n = 44k + r
Now count positive integers less than 100.
Case 1: r = 0
Numbers: 44, 88
-----> 2 numbers
Case 2: r = 1
Numbers: 1, 45, 89
-----> 3 numbers
Case 3: r = 2
Numbers: 2, 46, 90
-----> 3 numbers
Case 4: r = 3
Numbers: 3, 47, 91
-----> 3 numbers
Total = 2 + 3 + 3 + 3 = 11
Answer: D
Since the remainder when dividing by 4 must be less than 4,
r = 0, 1, 2, or 3.
A number with remainder r when divided by both 4 and 11 satisfies:
n ≡ r (mod 4)
n ≡ r (mod 11)
This means:
n − r is divisible by both 4 and 11.
Since LCM(4,11) = 44,
n = 44k + r
Now count positive integers less than 100.
Case 1: r = 0
Numbers: 44, 88
-----> 2 numbers
Case 2: r = 1
Numbers: 1, 45, 89
-----> 3 numbers
Case 3: r = 2
Numbers: 2, 46, 90
-----> 3 numbers
Case 4: r = 3
Numbers: 3, 47, 91
-----> 3 numbers
Total = 2 + 3 + 3 + 3 = 11
Answer: D
13 Jul 2026, 22:30
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The common remainder r can be 0, 1, 2, 3 .
we need to find,
n = 44 k + r
where r = 0, 1, 2, 3
now positive integer, n < 100
r = 0 n = 44, 88
r = 1 n = 1, 45, 89
r = 2 n = 2, 46, 90
r = 3 n = 3, 47, 91
So, total count of values of n is 11 (44, 88, 1, 45, 89, 2, 46, 90, 3, 47, 91)
Answer : D (11)
we need to find,
n = 44 k + r
where r = 0, 1, 2, 3
now positive integer, n < 100
r = 0 n = 44, 88
r = 1 n = 1, 45, 89
r = 2 n = 2, 46, 90
r = 3 n = 3, 47, 91
So, total count of values of n is 11 (44, 88, 1, 45, 89, 2, 46, 90, 3, 47, 91)
Answer : D (11)
Bunuel
13 Jul 2026, 23:24
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remainders that will be same are 0,1,2,3
since it is given as a positive integers, the numbers start from 1, 2, 3 = all the have the same remainders when divided by both 4 and 11
lcm of 4,11 is 44
44, 45, 46, 47 all these when div by both 4 and 11 will give the remainders 0,1,2,3 respectively
the next sync is 88, which is also under 100
so 88, 89, 90, 91 all these when div by both 4 and 11 will give the same remainders 0,1,2,3 respectively
the next sync will be greater than 100
so the number of positive integers less than 100 that have the same remainder upon division by 4 is 11 (1,2,3,44,45,46,47,88,89,90,91)
thanks,
Swetha
since it is given as a positive integers, the numbers start from 1, 2, 3 = all the have the same remainders when divided by both 4 and 11
lcm of 4,11 is 44
44, 45, 46, 47 all these when div by both 4 and 11 will give the remainders 0,1,2,3 respectively
the next sync is 88, which is also under 100
so 88, 89, 90, 91 all these when div by both 4 and 11 will give the same remainders 0,1,2,3 respectively
the next sync will be greater than 100
so the number of positive integers less than 100 that have the same remainder upon division by 4 is 11 (1,2,3,44,45,46,47,88,89,90,91)
thanks,
Swetha
Bunuel
