The ratio of cupcakes to children at a particular birthday

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7 children eat 7x cupcakes.Let Z be the no. of cupcakes left, less than 7
7x+ z=104 7x=104-z. Now, 104-z must be more than 97 and must be divisible by 7, it can be only 98
7x=98+6 6 cupcakes left, so it is a multiplier of 2 and 3.

What is being asked is what is the remainder and its factors.

Given that 104 is not a multiple of 7 AND given that the proportionality of the ratio will be kept (unless otherwise stated), no matter which (f) factor you apply:
You can express the ratio as a division with remainder (it is stated in the problem that R(f)<7(f)) to account for all the cupcacakes:

\(104(f)=7x(f) + R(f)\)
\(104(f)=7(14)(f) + 6(f)\)

The remainder is 6(f), so it must be a multiple of 2 and 3, no matter by which (f) the R (and the entire ratio) is multiplied. You can substitute the (f) for other numbers and 6 and (f) are, obviously, the factors of the remainder.

0<=104y-7xy<7y
0<=104-7x<7
13 6/7<x<=14 6/7
x=14

Uneaten:
104-7(14)=6 (mult. of 3 and 2, so D)

Is it also correct to just do it like this:

Assuming that each kid eats the exact same amount of cupcakes just pick any number. So any multiple of 7 can be taken away from 104, i took 14 (each kid eats 2). 90 is only divisible by 2 and 3 not by seven.

One additional condition is that number of leftover cupcakes must be less than the number of kids. So 90 leftover cupcakes is not correct. But even if we ignore this condition, note that 90 is divisible by 2 and 3 but not by 7 so we can say that 7 is certainly out. But can we say that in every case, the leftover cupcakes WILL BE divisible by 2 and 3? Not necessary!

Check my solution above which assumes values but uses logic at the end to ensure that it will hold in every case and hence must be true: the-ratio-of-cupcakes-to-children-at-a-particular-birthday-143833.html#p1332088

I still don't understand why it can't also be a multiple of 7? And hence why the answer can't be E instead of D. Please explain?

Check this first: the-ratio-of-cupcakes-to-children-at-a-particular-birthday-143833.html#p1152610

The question asks what MUST be true about the number of uneaten cupcakes? Not COULD be true. Thus, while the number of uneaten cupcakes COULD be a multiple of 7 (for example if k=7), it's not necessary to be (for example if k=1).

Hope it's clear.

Another approach is to use numbers and reasoning.

The less number to create the ratio

# Cupcakes (C)=104
# Children or kids (K) =7

C/k =104/7

After each kid has eaten x cupcakes the number left was less than # of kids. It means leftover (uneaten) is less than 7. It is between 1-6 cupcake.

Based on above we can eliminate choices C & E. 1 to 6 can't be multiple of 7

Now we need to work backward to find the number eaten by kids and the number should divide into 7. With sense of numbers 105 divides into 7. The next number divides into 7 is 98. We do not need to continue as uneaten cupcakes not huge.

Let uneaten cupcakes = 6
104 - 6 =98 ...........Hence the whole kids ate 98 cupcakes then each kid must have eaten x= 14 cupcakes

104- 5= 99.............x can't divide into 7

No need to continue till 1 as there is no number divides into 7 between 98 & 105

Hence uneaten cupcakes (6) should multiple of 2 & 3

Answer: D

let x=14 cupcakes
7*14=98 cupcakes eaten
104-98=6 cupcakes uneaten
I and II only
D

cupcakes/children = 104/7
given that
the number of cupcakes that remain uneaten is less than the number of children at the birthday party
max cupcakes that can be eaten has to be multiple of 7 <104 ; i.e 98 7*14
so left cupcakes = 6
which is multiple of 2 & 3
IMO D

Isn't there a simple logic behind it, that the number of uneaten cupcakes is less then the number of children and the number of children will be a multiple of 7, hence it can be 7 too but can not be less than 7 since there is no possibility of having 0 children in the party. Therefore, the number of uneaten cupcakes can never be a multiple of 7 and can only be multiples of 2 or 3 or 5.