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

1 2 3 4 5

Rishm

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13 Jul 2026, 12:22

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E is the answer.

I basically saw what has 0 as reminder which is 44 and 88, now look for +1,+2,+3 numbers, you will see - 45,46,47,89,90,91 as numbers, adding 1,2,3 as well as they give the same reminder. Hence 12 is the correct answer here.

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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

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Integer will have same remainders when divided by 4 and 11
on dividing by 4 we can get remainders - 0,1,2 and 3
we can say the same as the remainders when the number will be divided by 11

we take 4 cases
Case 1 - Remainder = 0
Here we can simply take Commom multiple of 4 and 11 before 100 and we get 44 and 88
Hence 2 Numbers

Case 2- Remainder = 1
We can denote the series as 4a + 1 and 11b + 1
4a + 1 = 11b + 1
4a = 11b
As this cannot be factorised we can say a = 11 and b = 4
We can conclude that every 11th term in 1st series (4a + 1) will be the same as every 4th term in 2nd series (11b + 1)
Based on this we get the integers 45 and 89 as the integers with remainders = 1 when dividing by 4 or 11
Hence 2 numbers

repeating the same steps for
Case 3 - remainder = 2
Common terms = 46 and 90

and Case 4 - remainder = 3
Common terms = 47 and 91

Hence Answer is (A) 8 terms

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13 Jul 2026, 12:27

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Let K and Q be intergers,

according to question 4K + x = 11Q + x

K/Q=11/4

Let Q = 4, it implies K=11 This 44,45,46,47
Let Q = 8, it implies K =22 This 88,87,98,90
Let Q = 0, it implies K = 0 This 0,1,2,3 (cant be 0 as the question says +ve integers)

total number of cases = 4+4+3 = 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

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1,2,3,44,45,46,47,88,89,90,91. Total 11 nimbers

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LCM(4,11)=44
n=44k+r
r can be 0,1,2,3
k=0, r=1,2,3..n=1,2,3
k=1, r=0,1,2,3...n=44,45,46,47
k=2, r=0,1,2,3...n=88,89,90,91

total 11 nos.

Ans 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

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n = 4x + r
n = 11y + r

n = 44z + r

r can be 0, 1, 2, or 3

When r = 0 ; n = 44, 88
r = 1 ; n = 1, 45, 89
r = 2 ; n = 2, 46, 90
r = 3 ; n = 3, 47 , 91

Number of integers = 9 + 2 = 11

Option D

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lets suppose, number of total integer = x
and each time the remainder = y

so, x ( mod 4) = y
x( mod 11) =y

LCM of 4 and 11 is 4, so , x(mod44)=y
if modulo is 4 remainder must be one of 0/1/2/3

so , suppose , x==44a+y where a = 0,1,2,3,4,.............

if a = 0 , x= 0+ ( 0,1,2,3)=(0,1,2,3)
if a = 1 , x= 44 +(0,1,2,3) =( 44,45 ,46 47)
if a = 2 , x= 88+(0,1,2,3)=(88,89,90,91)
if a =3 , x=132+(0,1,2,3)= not valid because , x must be greater than 0 and less than 100

so, possible value = 1,2 ,3 ,44 ,45,46,47,88,89,90,91

total possible value is 11 .

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My ans is C) 10

remainder when n is divided by 4 can be 0,1,2,3.

when remainder for both are same-
for n<4, remainder=n. Positive Integers = 1,2,3 will give remainders=integers upon division by both 4 and 11
1)0= 44,11
2)1= 45,89
3)2= 47,79
4)3= 91

Total count= 10. There can be a more efficient way for sure.

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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Answer:

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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

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When we divide a number by 4 the remainder can be 0, 1, 2 or 3.

This is because the largest value for the remainder is 3, which's one less than 4.

We can write the number as n = 44k.

The remainder, where 44 is the product of 4 and 11.

To find the values of n we need to look at the valid values of the remainder for numbers that are less, than 100.

If the remainder is 0 the numbers would be 44, 88 which gives us 2 numbers.
If the remainder is 1 the numbers would be 1, 45 89 which gives us 3 numbers.
If the remainder is 2 the numbers would be 2, 46 90 which gives us 3 numbers.
If the remainder is 3 the numbers would be 3, 47, 91 which gives us 3 numbers.

Total 11 numbers

IMO Option D

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Answer: D = 11

Step 1. LCM of 4 and 11 = 44

Step 2. Possible remainders 0 - 3 as 4 is the limiting integer

Step 3. Remainder (R) = 0

((Highest multiple of 44 - lowest multiple of 44) / 44 ) + 1

((88-44)/44) + 1 >> 2

0 is not included as lowest multiple of 44 since it is neither positive or negative and our set only includes positive integer less than 100

Step 4. R = 1

(( (88+1) - (1)) / 44) +1 >> 3

Step 5. R = 2 and 3

Same pattern holds as the highest multiple doesn't exceed 100, so 3 and 3 respectively

Total: 2(R=0) + 3(R=1) + 3(R=2) + 3(R=3) = 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

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4 and 11 have shared remainders=> 0,1,2,3

values for 44x + R, When remainder is 0 => 44, 88
when remainder is 1 => 1, 45, 89
when remainder is 2 => 2,46,90
when remainder is 3=> 3,47,91

total 11 positive integers less than 100.

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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13 Jul 2026, 17:43

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When we divide anything by 4, the reminder will be 0, 1, 2, 3
Therefore, when we divide anything by 11, the reminder must be 0, 1, 2, 3

Before finding the list, lets do an LCM of 4 and 11. It will be 44. Then, there will be no reminder.
Here is the list of positive integers less than 100
Reminder 0: 44, 88 (88/4 = 11) = 2
Reminder 1: 1, 45, 89 (11*4, reminder 1, 11*8, and 4*22, reminder 2) = 3
Reminder 2: 2, 46, 90 [11*4, reminder 2, 11*8, and 4*22, reminder 2] = 3
Reminder 3: 3, 47, 91 (11*4, reminder 3, 11*8, and 4*22, reminder 3) = 3

Total reminders =2+3+3+3 = 11

Answer: D

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By brute force pattern
Any number less than both 4 and 11 will leave the same reminder:1,2 and 3
A number which is factor of both 4 and 11 will leave the same reminder:0: Possible nums:44,88
And when we add the first set of numbers to the multiples of 4 and 11 then also the same remainders are left:
SO the possible set of numbers:1,2,3,44,45,46,47,88,89,90,91=>11 nums possible

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N = 4a+r
N = 11b+r
equating we get a/b = 11/4;
Assume a = 11k, b = 4k
SO we get N = 44k + r
Now r can be 0,1,2,3 since it is has to leave same remainder as that when divided by 11
If r = 0
N = 44k so N can be 44,88
If r = 1
N = 44k+1 so N can be 1, 45, 89
similarly r=2 and r=3 give. 3 values each
So total values = 2+3+3+3 = 11 -> D

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13 Jul 2026, 19:56

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Question asked: positive integers < 100 have the SAME REMAINDER upon division 4 and 11

1. Upon Division 4 and 11 under 100 with remainder 0 -> 4x11 = 44, 88 (2)
2. with remainder 1: 1, 45, 89 (3)
3. with remainder 2: 2, 46, 90 (3)
4. with remainder 3: 3, 47, 91 (3)

1,2,3 are included since if divided by 4, will have reminder as above as well.

So, total 2+3+3+3 = 11 (Answer D).

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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D 11.

Need n mod 4 = n mod 11 = r. Since r must be a valid mod-4 remainder, r is 0, 1, 2, or 3. Then n minus r is divisble by both 4 and 11, so by 44. Thus n = 44k + r

List 1 to 99

k = 0, 1, 2, 3 (drop 0, it's not positive) -> three
k = 1: 44, 45, 46, 47 -> four
k = 2: 88, 89, 90, 91 -> four

total 11.

Wrong ones
E - 12: Counted k=0, as four, forgetting that it's not positive
C - 10: Dropped a boundary value either 88 or 91, both under 100
B - 9: Kept the two full blocks, but missed the small three (1, 2, 3)
A - 8: Undercounted repeats across blocks.

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13 Jul 2026, 21:19

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Lets check small numbers like 1,2 and 3 (less than 4) when 1 is divided by 4 or 11 reminder is 1 , same is the case with 2 and 3 when divided by 11 or 4 leaves a reminder 2, 3
now if we check for larger number like 32 we see the reminder is 0 when divided by 11 the reminder is 10 so for 2 digit numbers less than 100 we need to find first the common multiples of 4 and 11
44, and 88 are the only common multiples .
Check
44 when divided by 4 / 11 leaves a reminder 0
44+1=45 when divided by 4/11 reminder is 1
44+2 =46 when divided by 4/11 reminder is 2
44+3 =47 when divided by 4/11 reminder is 3
44+4 =48 when divided by 4 leaves reminder 0 when divided by 11 leaves reminder 4 . STOP. So - 1,2,3 ,44,45,46 .47 when divided by 4 or 11 leaves same reminder

No check for 88

88 when divided by 4/11 leaves reminder 0
88+1=89 when divided by 4/11 leaves reminder 1
88+2=90 when divided by 4/11 leaves reminder 2
88+3 =91 when divided by 4/11 leaves reminder 3
88+4 =92 when divided by 4/11 leaves reminder 0 for 4 and reminder 4 for 11 . STOP!

So 1,2,3,44,45,46,47 ,88,89,90,91 leaves same reminder for 4 and 11
no of terms = 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

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For 4 we can only have these remainders: 0,1,2,3
Let us consider the numbers when Remainder will be 0 : 44, 11
Remainder 1: 45,85
Remainder 2: 46, 86
Remainder 3: 47, 87

Apart from these 8 numbers, all other numbers have different remainders.
Eg: consider 34 => Remainder for 11 = 1
Remainder for 4 = 2.

ANSWER A

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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

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OPTION D. 4 and 11 are co prime. from trial and error we know 22 gives remainder 0 with 11 and 2 with 4. for 33 its 0 and 1. then 44, both gives 0, as for 4 only 1,2,3 are possible remainders. 44,45,46,47 are nos. similarly 88,89,90,91. therefore 4+4 + 3(initial numbers 1,2,3) = 11