How many numbers lie between 11 and 1111 which when divided by 9 leave

Question

How many numbers lie between 11 and 1111 which when divided by 9 leave a remainder of 6 and when divided by 21 leave a remainder of 12?

A.16
B.17
C.18
D.19
E.20

pradhan · Accepted Answer

Oh!
My answer is C. 18

It is but natural that the numbers we are interested in are separated by 63 LCM of 9 and 21.

The first number is 33 and all others are separated by 63. It forms an arithmatic progression.

I am using the nth term formula of AP
nth term = a+ (n-1)d where a=33 and d=63.
the requirement is nth term less than 1111
hence a+(n-1)d<1111
33+(n-1)*63 < 1111
n-1<1078/63
n-1<17.11
n<18.11
the 17th term as per the formula a+(n-1)d is 1041
and the 18th term is 1104

kalpeshchopada7 · Answer

I solved it little differently. I took the first no. by multplying 21 by 1 and adding 12. i get the result 33 which fits the second test as well. then, 21 by 2 add 12 does not fit the test ans so do 21 by 3 + 12. However, 21 by 4 +12 does. that means the number has to be in the sequence with 1,4,7,10... 52 multiples of 21 as the number beyond 52 is outside the given set of numbers.

52+1/3 must be >17 hence the answer should be 18.

52+1/3 must be >17 hence the answer should be 18.

amit2k9 · Answer

lcm = 63
a = 33

a+(n-1) d < 1111

giving n = 18

subhashghosh · Answer

N = 9K + 6 = 6,15,24,33,

N = 21m + 12 = 12, 33..

=> N = 63n + 33

Max value of N < 1111 = 63* 17 + 33

So total # = 18

Because n = 0 is the first number's index.

Answer = C

KarishmaB · Answer

Check this link for more on how to solve such questions: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... s-part-ii/

JeffTargetTestPrep · Answer

The first number that leaves a remainder of 6 when divided by 9 is 6.

Then we have 6, 15, 24, 33, 42, 51, 60, and so on.

The first number that leaves a remainder of 12 when divided by 21 is 12.

Then we have 12, 33, 54 and so on.

We can see that 33 is the smallest number that satisfies both conditions.

Since the LCM of 9 and 21 is 63, we can keep adding 63 to 33 to find the succeeding numbers that satisfy both properties:

96, 159, ...

While we could list the numbers until we reach but not exceed 1111, a more efficient way is to find the find the largest number less than 1111 that satisfies both conditions, by the following equation:

33 + 63n < 1111

63n < 1078

n < 17.11

Since n is an integer, n = 17. That is, 33 + 63(17) is the largest integer that satisfies both conditions. Since 33 + 63(0) = 33 is the smallest integer that satisfies both conditions, we have 18 integers (in the form of 33 + 63m, where m is an integer from 0 to 17, inclusive) that satisfies both conditions.

Answer: C

WiziusCareers1 · Answer

Let the number be N. We are given two conditions:
N = 9k + 6 (Remainder is 6 when divided by 9)
N = 21m + 12 (Remainder is 12 when divided by 21)

We can list values for each to find the first common N:
Values for 9k + 6: 6, 15, 24, 33, 42...
Values for 21m + 12: 12, 33, 54...
The smallest number that satisfies both is 33. This will be our first term (a).

The numbers that satisfy both conditions will repeat every LCM(9, 21), which is 63
So, the numbers form an Arithmetic Progression (AP) where the first term a = 33 and the common difference d = 63.

We need to find how many such numbers N lie between 11 and 1111. The general formula for the nth
term of an AP is a+(n−1)d
We set the inequality 
33 + (n − 1)63 < 1111
(n − 1)63 < 1078
n − 1< 1078/63
n - 1 < 17. 11
n < 18.11
Since n must be an integer, the largest value for n is 18.

Hence, the answer is C.

Adit_ · Answer

Remainders of the form 21x+6 when considered keep giving remdinder of 6 1/3 times in cyclical form when divided by 9.

21x=1111 
x = 52.

Total 52 such values now we need to find total numbers where it satisfies both conditions.

If we see first value 21*1+12 = 33 = 9*6.
Thus every 1,4,7,10..give 6 as remainder with 9.

Using a+(n-1)d = 1+(n-1)3 = 52.
n = 18.

Answer:  Option C

egmat · Answer

Let's break this down step by step.

We need numbers N between  11  and  1111  such that:
- N divided by  9  leaves remainder  6  → N =  9 a +  6 
- N divided by  21  leaves remainder  12  → N =  21 b +  12

Step 1:  Start with the second condition.
N =  21 b +  12  for some whole number b.

Step 2:  Plug into the first condition.
We need  21 b +  12  to leave remainder  6  when divided by  9 .
 21  =  9  ×  2  +  3 , so the remainder from  21 b is  3 b. Since  12  =  9  ×  1  +  3 , the remainder from  12  is  3 .
So we need  3 b +  3  to give remainder  6  when divided by  9 .
This means  3 b +  3  =  6 , or  15 , or  24 ... In other words,  3 b must equal  3  (mod  9 ), so b must equal  1  (mod  3 ).

Step 3:  Write b in terms of a new variable.
b =  3 k +  1 , where k =  0 ,  1 ,  2 ,  3 ...

Substitute back: N =  21 ( 3 k +  1 ) +  12  =  63 k +  21  +  12  =  63 k +  33

Key Insight:  Every valid number follows the pattern N =  63 k +  33 .

Step 4:  Find how many values of k give N between  11  and  1111 .
- Lower bound:  63 k +  33  >  11  → k ≥  0 
- Upper bound:  63 k +  33  <  1111  →  63 k <  1078  → k <  17.11

So k can be  0 ,  1 ,  2 , ...,  17  — that's  18  values.

Quick check: k =  0  gives  33  ✓ (between  11  and  1111 ), k =  17  gives  1104  ✓.

Answer: C (18)

How many numbers lie between 11 and 1111 which when divided by 9 leave

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