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GMAT Club World Cup 2026 (Day 6): How many positive integers less than

1 2 3 4 5

Bunuel

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How many positive integers less than 100 have the same remainder upon division by 4 as upon division by 11?

A. 8
B. 9
C. 10
D. 11
E. 12

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Bunuel

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GMAT Club Official Solution:

How many positive integers less than 100 have the same remainder upon division by 4 as upon division by 11?

A. 8
B. 9
C. 10
D. 11
E. 12

The common remainder must be 0, 1, 2, or 3, since these are the only possible remainders when a number is divided by 4.

If a positive integer has remainder r when divided by both 4 and 11, then after subtracting r, the result must be divisible by both 4 and 11, so it must be divisible by the LCM of 4 and 11, which is 44. Thus, the number must be of the form:

44k + r, where r = 0, 1, 2, or 3.

Now count the positive integers less than 100.

• For k = 0, we get:

1, 2, 3
(discard 0 because it is not a positive integer)

• For k = 1, we get:

44, 45, 46, 47

• For k = 2, we get:

88, 89, 90, 91

Total:

3 + 4 + 4 = 11

Answer: D.

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Here's my solution for this question

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When any number is divided by 4, the remainders can be only 0,1,2,3.
LCM (4,11)= 44
Now add possible remainders of 4 to LCM(4,11) to find other numbers that leave same remainder.
Multiples of 44 = 0, 44,88 (Remainder 0)
Add 1= 1, 45, 89 (Remainder 1)
Add 2= 2, 46, 90 (Remainder 2)
Add 3= 3, 47, 91 (Remainder 3)

Total numbers less than 100 that leave same remainder when divided by 4,11 = 2+3+3+3=11
0 is neither negative nor positive. So excluded.

D. 11

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13 Jul 2026, 08:27

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How many positive integers less than 100 have the same remainder upon division by 4 as upon division by 11?

A. 8
B. 9
C. 10
D. 11
E. 12

least remainder values will be 1,2,3

multiple of both 4 & 11
44,88
values ,45,46,47 , 89,90,91

total values where remainder is same when divided by both 4 & 11
1,2,3,44,45,46,47,88,89,90,91

total such values is 11
OPTION D is correct

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any number divided by 11 will give remainder from 0 to 10. And any number divided by 4 will give remainder as 0 to 3. Thus for the remainder to be same , number must be divisble by both and first two digit number divisble by both is 44 giving remainder as 0 then 45 gives remainder 1 in both case, similarly 46 & 47.
Now next number divisible by both is 88, similarly 89,90& 91 will give same remainder when divided by both.
Thus total of 8 numbers is ANS.

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r be reminder
r < 4
r = 0,1,2 or 3

LCM of 4 and 11 is 44
hence number n = 44x+r
number n <100
so for
r = 0, n = 44,88
r = 1, n = 1, 45,89
r = 2, n = 2, 46,90
r = 3, n = 3, 47, 91
total r = 2+3+3+3
=11
Option: D

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i am going with option d.
only possible remainders - 0, 1, 2 or 3.
matching numbers repeat every 44 units (lcm of 4 and 11), n = 44k+r
k=0: 3 numbers (1,2,3 - 0 is excluded as it is not positive)
k=1: 4 numbers (44,45,46,47)
k=2: 4 numbers (88,89,90,91)
total = 3+4+4 = 11
a and b incorrect - miss out entire blocks of valid numbers (1,2,3)
c incorrect - it occurs if forgets to count (1,2,3) or mistakenly add 0.
e incorrect - it occurs if you incorrectly count 0 as valid number or extend pattern past 100 (132 - quite high)

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n = 4k + r = 11m + r < 100
4k = 11m
k = 11m/4; m is a multiple of 4

11m + r < 100
m < (100-r)/11; m<10; m={0,4,8}

m = 0; n = {1,2,3}; 0 is excluded since it is NOT a positive integer
m = 4; n = {44,45,46,47}
m = 8; n = {88,89,90,91}

n = {1,2,3,44,45,46,47,88,89,90,91}
11 positive integers less than 100 which have the same remainder upon division by 4 as upon division by 11.

IMO D

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postive interger less than 100 leaving same remainder when divided by 4 and 11
number <4 = 1,2 and 3 leave remainder leave same remainder when divided by 4 n 11

lcm of 4 and 11 = 44
44+
45, 46, 47 leave remainder of 1,2,3 respectivelly
88+
89,90 and 91
in all 9 nos option A

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Divided by 4 remainder can only be 0,1,2,3 (we can think of 11 also but it has 10 remainders so many more cases, 4 is easier)

For 0:

44 / 88 both give 0 remainder when divided w 11 or 4 = 2

For 1 : 1 / 45 / 89 = 3

For 2: 2 / 46 / 90 = 3

For 3: 3 / 47 / 91 = 3

11

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Answer: D) 11

x/4 = 4C + r and x/11 = 11C + r

If n is divisible by both 4 and 11 when there is no remainder (i.e. n - r) then n must be a multiple of LCM(4,11) plus said remainder.

n = 44C + r

Because we're looking for the values of n less than 100 we only need to test the values of C = 0,1,2

n = 44(0) + r
n = r
r can only take 0,1,2 or 3 as possible values because any more than that and you would be able to fit one 4 in the remainder.
The Question also asks for all positive integers so we can only consider 1, 2 or 3 -> three possible values

n = 44(1) + r
n = 44 + r, n = 44, 45, 46, or 47 -> four possible values

n = 44(2) + r
n = 88 + r, n = 88, 89, 90, or 91 -> four possible values

3 + 4 + 4 = 11 possible values.

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To find numbers from 1 to 99 that gives same remainder when divided by 4 and 11.
Division by 4 can only leave 0,1,2,3 as remainders.
Listing numbers which, when divided by 4 and 11 gives following remainders:
0 : 44, 88
1 : 1, 45, 89
2 : 2, 46, 90
3 : 3, 47, 91

In total we have 2+3+3+3 = 11 numbers which leave same remainders when divided by 4 and 11

Bunuel

How many positive integers less than 100 have the same remainder upon division by 4 as upon division by 11?

A. 8
B. 9
C. 10
D. 11
E. 12

This question was provided by GMAT Club
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assuming common remainder=r
since remainder mod 4 is <4
r= 0,1,2,3
n--r (mod 44)
less than 100
r=0 ; 44,88--2
r=1; 1,45,89--3
r=2,46,,90--3
r=3,47,91--3

2+3+3+3=11

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for the numbers to have same remainders, it should be divided by 44 (11*4)
and the remainders should be less than the smaller number <4
ie. Remainders can be either 0,1,2,3

i.e. 44x+R = N

Case 1 - R = 0 ->with X as 0 - we get n as 0, x as 1, n = 44, x=2, n = 88, x=3, n>100 [Count = 2, as n cannot be 0, as it is not a positive integer)
Case 2 - R = 1 -> with x as 0 - we get n as 1, x as 1, n = 45, x=2, n=89, x=3, n>100 [Count = 3]
Case 3 - R = 2 -> with x as 0, - we get n as 2, x as 1, n = 46, x =2, n = 90, x=3, n>100 [Count = 3]
Case 4, R = 3 -> X as 0, N = 3, x as 1, n = 47, x as 2, n = 91, x=3, n>100 [Count = 3]

Total Count = 2+3+3+3 = 11

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As the division is with 4 and 11, the remainders will be 0,1,2,3 and 0,1,2,3,....10 respectively for both of them. So only possible common remainders will be 0,1,2,3

4 and 11 are co-primes, i.e. they share only HCF of 1.

The LCM of 4 and 11 will be 44. So at 44 both of these will have remainder of 0. 4 and 11 will have the same remainders for 44,45,46,47 i.e., 0,1,2,3 ==> 4 integers

The next multiple of 44 is 88, so at 88 the remainder will be 0. 4 and 11 will have the same remainders for 88,89,90,91 i.e. 0,1,2,3. ==> 4 intergers

Also, while dividing both 4 & 11 by 1,2,3 we will have the same remainder of 1,2,3 ==> 3 integers

Total = 4+4+3 =11 Integers.

Answer(D)

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Positive integers means 1,2,3,...

they should have same remainder upon division by 4 as division by 11. Means when number is divided by 11, it will have 0,1,2,3 number as the remainder.
The number will follow 44a + r pattern. The whole number 44a+r has to be lower than 100.

when a=0, r could be 1,2,3 (positive integers only)
when a=1, r could be 0,1,2,3
when a=2, r could be 0,1,2,3
when a=2, number will be >100 and hence not allowed.

Total number of integers satisfying the criteria mentioned in the question stem is 3+4+4 = 11 nos.

Bunuel

How many positive integers less than 100 have the same remainder upon division by 4 as upon division by 11?

A. 8
B. 9
C. 10
D. 11
E. 12

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Updated on: 06 Aug 2026, 00:51

Originally posted by GulfTube on 13 Jul 2026, 09:25.
Last edited by GulfTube on 06 Aug 2026, 00:51, edited 1 time in total.

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let n be the positive integer less than 100
dividing by can give remainders : 0,1,2,3
dividing by 11 can give remainders: 0, 1,2,3,....,10

since we are interested in common remainder, we will focus on remainders of 0,1,2,3
n divided by 4, possible values of n are :
remainder 0: 4,8,12,....96
remainder 1: 1,5,9,...97
remainder 2: 2,6,10,.....98
remainder 3: 3,7,11,....99

n divide by 11, possible values of n are:
rem 0: 11,22,33,44,....99
rem 1: 1,12,23,....98
rem 2: 2,13,24,....90
rem 3: 3,14,25,....91

LCM of 4 & 11 = 44
for common rem of 0: n can be: 44, 88 = 2 possible values
for common remainder of 1: n can be: 1, 45, 89 = 3 possible values
for common rem of 2: n can be: 2, 46, 90 = 3 possible values
for common rem of 3: n can be: 3, 47, 91 = 3 possible values
total possible values of n : 11

D

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Bunuel

How many positive integers less than 100 have the same remainder upon division by 4 as upon division by 11?

A. 8
B. 9
C. 10
D. 11
E. 12

This question was provided by GMAT Club
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For any positive integer n, lets say the reminder is r.
when n is divided by 4, r should be less than 4
so, r = {0,1,2,3}
we can write,
when divided by 4,
n-r = multiple of 4
when divided by 11,
n-r = multiple of 11
since the LHS be multiple of both 4 and 11, then it should be a multiple of LCM(4,11) = 44
n-r = multiple of 44
so, n-r = 44k, for some non-negative integer k
lets check and verify
1. if k = 0
n = 0 +r, but we know that r = {0,1,2,3}
so, n = {0,1,2,3} since n is positive , n = {1,2,3} - 3 integers
2. if k = 1
n = 44+r, r = {0,1,2,3}
so, n = {44,45,46,47} - 4 integers
3. if k = 2
n = 88+r, r={0,1,2,3}
so, n={88,89,90,91} - 4 integers
4. if k = 3,
n = 132+r, r={0,1,2,3}
this implies n>100 - contradictory- so not counted

so no of values of n = 3+4+4 = 11

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Any number divided by 4 gives remainders 0, 1, 2 and 3
So any number dividing by 11 with the same remainder is an answer.

Only +ve numbers so 0 cannot be an option. But for 1 ,2 and 3 both 4 and 11 will have the same remainders. = > 3 numbers
similarly for 44. 44 gives 0 remainder for both, 45 gives 1, 46 gives 2 and 47 gives 3 before resetting to 0 for 4 = > 4 numbers
Same case for 88 => 4 numbers

for any other number, take 11 as example. 0 for 11, 1 for 4. the remainder mismatch will stay even if we keep increasing the number

Total numbers matching criterion 3 + 4 + 4 = 11. Answer D