Question

How many numbers lie between 11 and 111 which when divided by 9 leaves a remainder of 6 and when divided by 21 leaves remainder of 12?

Answer 1

Some methods :

1 ) 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.

2 ) 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

3 ) 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.

4 ) number lies between 11 and 1111

When divide by 9 a remainder of 6 and when divided by 21 leave a remainder of 12 = 33, 96, 159 (all separated by 63)

a =33  d=63

1111=a+(n-1)631111
=33+(n-1)631078
=(n-1)631078/63
=n-1n
=(1078+63)/63n
= 1141/63n
= 18.11

then total number=18

Answer is 18.

Hope This Helps :)

Answer 2

Answer:

The total $18$ numbers lie between $11$ and $111$ which when divided by $9$ leaves a remainder of $6$ and when divided by $21$ leaves remainder of $12.$

Step-by-step explanation:

Let the possible number be $N$ then it are often expressed as

$N = 9k + 6$

and $N=21I+12$

∴$9k + 6 = 21I + 12$

$9k - 21I = 6$

or $3(3k - 7I) = 6$

or $3k = 7I + 2$ or

$k=7I+23$

So put the minimum the possible value of $I.$such that the worth of $k$ is an integer or in other words numerator (i. e., $7I+2$) are going to be divisible by$3$.

Thus at $I=1$, we get $k=3$ (an integer) that the least possible number

$N = 9 \times 3 + 6 \\N= 21 \times 1 + 12 \\N= 33.$

Now the upper possible values is obtained by adding $33$ within the multiples of LCM of $9$and $21$. i.e., the final sort of the quantity is $63m + 33$. therefore the other number within the given range including $33$ are $96, 159, 222, 285, 348, ..., 1104.$

Hence there are total $18$ numbers which satisfy the given condition.

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