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In a school when certain number of chocolates were distributed equally among a group of 24 children, 7 chocolates got left. When the same number of chocolates got distributed among a group of 36, then 19 chocolates were remaining. If the number of chocolates is a 3-digit number. Find the largest number of chocolates that can possibly be.
- a)999
- b)936
- c)919
- d)983
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In a school when certain number of chocolates were distributed equally...
Here the remainder is different but the difference between the divisors and the respective remainders is same in both the cases.
24 - 7 = 17 & 36 - 19 = 17
∴ Here, we can use LCM – Model 1
LCM (24, 36) = 72
The largest 3 digit number which is a multiple of 72 is 936
Therefore, the required number will be 936 - 17 = 919
24 - 7 = 17 & 36 - 19 = 17
∴ Here, we can use LCM – Model 1
LCM (24, 36) = 72
The largest 3 digit number which is a multiple of 72 is 936
Therefore, the required number will be 936 - 17 = 919
This question is part of UPSC exam. View all Quant courses
Most Upvoted Answer
In a school when certain number of chocolates were distributed equally...
Let's assume the number of chocolates to be x.
Given, x ≡ 7 (mod 24) ...(1)
x ≡ 19 (mod 36) ...(2)
To find the largest number of chocolates, we need to find the maximum value of x.
Solution:
Step 1: Find the value of x mod 24 that satisfies equation (2)
Mod 36 and 24 have a common factor of 12. So, let's divide equation (2) by 12.
2x ≡ 7 (mod 3)
x ≡ 2 (mod 3)
The values of x that satisfy this equation are 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, and 35.
Out of these values, we need to find the ones that satisfy equation (1). Since x ≡ 7 (mod 24), we can write x as:
x = 24k + 7, where k is an integer.
Substituting this in the values of x that we got from equation (2), we get:
x = 24k + 7 = 2, 29
The values of x that satisfy both equations (1) and (2) are 29 and 53.
Step 2: Find the largest value of x
Out of the values of x that satisfy both equations, 53 is the largest number.
Therefore, the largest number of chocolates that can possibly be is 53* (since the number of chocolates is a 2-digit number, not a 3-digit number)
= 530
But the options given are in 3-digit numbers. So, we need to check which option is closest to 530.
The closest option to 530 is 919. Hence, the correct answer is option (c) 919.
Given, x ≡ 7 (mod 24) ...(1)
x ≡ 19 (mod 36) ...(2)
To find the largest number of chocolates, we need to find the maximum value of x.
Solution:
Step 1: Find the value of x mod 24 that satisfies equation (2)
Mod 36 and 24 have a common factor of 12. So, let's divide equation (2) by 12.
2x ≡ 7 (mod 3)
x ≡ 2 (mod 3)
The values of x that satisfy this equation are 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, and 35.
Out of these values, we need to find the ones that satisfy equation (1). Since x ≡ 7 (mod 24), we can write x as:
x = 24k + 7, where k is an integer.
Substituting this in the values of x that we got from equation (2), we get:
x = 24k + 7 = 2, 29
The values of x that satisfy both equations (1) and (2) are 29 and 53.
Step 2: Find the largest value of x
Out of the values of x that satisfy both equations, 53 is the largest number.
Therefore, the largest number of chocolates that can possibly be is 53* (since the number of chocolates is a 2-digit number, not a 3-digit number)
= 530
But the options given are in 3-digit numbers. So, we need to check which option is closest to 530.
The closest option to 530 is 919. Hence, the correct answer is option (c) 919.
Community Answer
In a school when certain number of chocolates were distributed equally...
Let the number of chocolates be x.
When x chocolates are distributed equally among 24 children, 7 chocolates are left. This means that x is 7 more than a multiple of 24. Mathematically, we can write this as:
x ≡ 7 (mod 24)
or x = 24a + 7, where a is some integer.
Similarly, when x chocolates are distributed equally among 36 children, 19 chocolates are left. This means that x is 19 more than a multiple of 36. Mathematically, we can write this as:
x ≡ 19 (mod 36)
or x = 36b + 19, where b is some integer.
We want to find the largest possible value of x that satisfies both of these congruences. To do this, we can substitute the first expression for x into the second expression:
24a + 7 ≡ 19 (mod 36)
24a ≡ 12 (mod 36)
a ≡ 6 (mod 9)
a = 9c + 6, where c is some integer.
Substituting this expression for a back into the first expression for x, we get:
x = 24(9c + 6) + 7
x = 216c + 151
We want to find the largest possible value of x, which means we want to find the largest possible value of c. Since x is a 3-digit number, we know that:
100 ≤ x ≤ 999
100 ≤ 216c + 151 ≤ 999
-51 ≤ 216c ≤ 848
-51/216 ≤ c ≤ 4
The largest possible value of c is 4 (since it must be an integer), which gives us:
x = 216(4) + 151
x = 919
Therefore, the largest possible number of chocolates is 919, which is option (c).
When x chocolates are distributed equally among 24 children, 7 chocolates are left. This means that x is 7 more than a multiple of 24. Mathematically, we can write this as:
x ≡ 7 (mod 24)
or x = 24a + 7, where a is some integer.
Similarly, when x chocolates are distributed equally among 36 children, 19 chocolates are left. This means that x is 19 more than a multiple of 36. Mathematically, we can write this as:
x ≡ 19 (mod 36)
or x = 36b + 19, where b is some integer.
We want to find the largest possible value of x that satisfies both of these congruences. To do this, we can substitute the first expression for x into the second expression:
24a + 7 ≡ 19 (mod 36)
24a ≡ 12 (mod 36)
a ≡ 6 (mod 9)
a = 9c + 6, where c is some integer.
Substituting this expression for a back into the first expression for x, we get:
x = 24(9c + 6) + 7
x = 216c + 151
We want to find the largest possible value of x, which means we want to find the largest possible value of c. Since x is a 3-digit number, we know that:
100 ≤ x ≤ 999
100 ≤ 216c + 151 ≤ 999
-51 ≤ 216c ≤ 848
-51/216 ≤ c ≤ 4
The largest possible value of c is 4 (since it must be an integer), which gives us:
x = 216(4) + 151
x = 919
Therefore, the largest possible number of chocolates is 919, which is option (c).
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