A woman who has two sons enrolled in an elementary school brings a bat

We can form two equations from the given information in the question.
11a+5=n..........(1)
8b+7=n............(2)
where a,b, and n are integers with n more than 200.
LCM(8,11)=88
We need to determine the least value of n which satisfy both equations above.
Equating both equations, we get 11a+5=8b+7
hence 11a-8b=2
we need a multiple of 11 which differs from the multiple of 8 by 2. Definitely that number will be an even multiple of 11. If a =6, we get 11*6=66 and this is 2 more than 64. Hence the first value of n occurs when a=6 or b=8.
n=66+5=64+7=71.
The rest of the task is just to keep adding multiples of 88 to 71 until we get the first number which is more than 200.
n=71+88=159+88=247
Hence n is 247.
Sum of the digits of n=2+4+7=13.

The answer is therefore C.

Elder son: n =7+8a ={7,15,...,} 207,215,...
Younger son: n =5+11a ={5,16,...,} 203,214,...

Try to match the possible N between the two sons:
Elder: 207, 215, 223, 231, 239, 247
Younger: 203, 214, 225, 236, 247

So the sum of the digits of the smallest possible value of n is 2+4+7=13

FINAL ANSWER IS (C)

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youngest+10=11: n/11=11p+5=[5,16,27…71…]
oldest+7=8: n/8=8q+7=[7,15,23,…71…]
n>200: n=lcm(11,8)k+71=88k+71=88*2+71=176+71=247
min(n)=247; sum.digits(n)=2+4+7=13

Ans (C)

A woman who has two sons enrolled in an elementary school brings a batch of n cookies to the school. Her youngest son is in a class with 10 other students and her oldest son is in a class with 7 other students. If she divides the n cookies evenly among the students in her youngest son’s class there will be 5 left over. If she divides the cookies evenly among the students in her oldest son’s class there will be 7 left over. If she has more than 200 cookies, which of the following is the sum of the digits of the smallest possible value of n?

A. 9
B. 11
C. 13
D. 17
E. 23

n > 200 and n = 11k + 5 OR n = 8k' + 7
Here 200 = 8 * 25 and since 200 = 11*18 + 2, minimum value of n where n/11 leaves a remainder of 5 should be such that value above 200 should leave a remainder of 3 satisfying the condition that n/8 leaves a remainder of 7.

Lowest such value is 247 which satisfies the condition.

Hence sum of digits is = 2 + 4 + 7 = 13.

Answer C.

Youngest son's group - 8 students (including himself)
Oldest son's group - 11 students (including himself)

n= 11a +5
n=8b +7

n >200
which of the following is the sum of the digits of the smallest possible value of n?
-------------------
11a +5 =8b +7
11a = 8b+2
--> "11a" have to be greater than 200 and be even number

if 11a= 220, then --> \(b= \frac{218}{8}\) (not integer)

if 11a = 242, then --> \(b=\frac{240}{8}= 30\)

--> n=11a+ 5= 242+5= 247
the sum of digits of 247 - (2+4+7) = 13

--> The answer is C

for youngest student class ; n/10 = 5
for eldest son class ; n/7 =7
n>200
n=11a+5
n=8b+7
least value is 71 ; and for multiple of 71 LCM = 88 ; values ; 71+88 159; 247

least value of n = 247 ;
IMO C ; 13 " sum of digits"


A woman who has two sons enrolled in an elementary school brings a batch of n cookies to the school. Her youngest son is in a class with 10 other students and her oldest son is in a class with 7 other students. If she divides the n cookies evenly among the students in her youngest son’s class there will be 5 left over. If she divides the cookies evenly among the students in her oldest son’s class there will be 7 left over. If she has more than 200 cookies, which of the following is the sum of the digits of the smallest possible value of n?

A. 9
B. 11
C. 13
D. 17
E. 23

The younger son's class has total 11 students and older son's class has total 8 students.
n = 11A + 5
n = 8B + 7 (this tells us that n is odd. So A must be even)

Let's find the first such possible value of n by hit and trial.
If A = 2, n = 27 (gives remainder 3 upon division by 8 - Not valid)
If A = 4, n = 49 (gives remainder 1 upon division by 8 - Not valid)
If A = 6, n = 71 ((gives remainder 7 upon division by 8 - VALID)
So the first value of n is 71.

All other values will be of the format n = 88M + 71
First value greater than 200 will be 88*2 + 71 = 247
Sum fo digits = 2 + 4+ 7 = 13

Answer (C)

We see that there are 11 students in her younger son’s class and 8 students in her older son’s class. We are given than n is greater than 200. Furthermore, we see that when n is divided by 11, the remainder is 5 and when n is divided by 8, the remainder is 7.

Even though we know that n is greater than 200, let’s start by listing the numbers (a lot smaller than 200) that leave a remainder of 5 when divided by 11:

5, 16, 27, 38, 49, 60, 71, …

We see that 71 is the smallest number that leaves a remainder of 5 when divided by 11 and also a remainder of 7 when divided by 8 (notice that 71/11 = 6 R 5 and 71/8 = 8 R 7). Now, to get the smallest value of n, we can keep adding the LCM of 11 and 8 (i.e., 88) to 71 until it becomes more than 200:

71 + 88 = 159

159 + 88 = 247

Therefore, the smallest value of n is 247, and the sum of the digits of 247 is 2 + 4 + 7 = 13.

Answer: C

Given, n>200.
Nos of students in young son class: 11
Therefore, n=11x+5(where x= nos of cookies evenly distributed among young sons class)
There 11x+5>200, x>17, x can be 18,19 and soon.
Hence N can be 203,214,225,236,247....

Elder son class:8
n=8y+7(where y= nos of cookies evenly distributed among elder sons class)
For 8y+7>200, y>24, y can be25,26,27...... Hence N can be 207,215,223,231,239,247

Therefore lowest possible value of N is 247.
Sum of digits:2+4+7=13
Answer: C

11a+5=8b=7
when b=8, a=6.
minimum number is 71.
Next number is LCM(11,8)=71
88k+71
when k=2, value is 176+71= 247
sum of the digits=2+4+7=13
Answer:C

Can be done fast by this approach:

Younger students class 11 students so total cookies in form of 11X + 5 (multiple of 11 plus 5).
multiple of 11 nearest to 200(total cookies are >200) is 198, 195 + 5 is 203 is cookies can be: 203, 214 (203+11), 225(214+11) and so on

same for older students class, cookies can be in form of 8Y+5

so can be 205, 213, 221 and so on

If you calculate, 247 is the common value for both

sum of digits of 247 is 13, hence answer is C.