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62 candies were equally distributed to a group of children. If the num

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62 candies were equally distributed to a group of children. If the num  [#permalink]

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New post 16 Oct 2018, 13:07
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A
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D
E

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62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15
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62 candies were equally distributed to a group of children. If the num  [#permalink]

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New post 16 Oct 2018, 14:31
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15


Let the number of children be x
the number of candies be c

the remainder equation can be set up as \(\frac{62}{x} = c + (\frac{x-2}{x})\)

Multiply by x

the equation becomes \(62 = c*x + x - 2\)

take x as a common factor it becomes \(62 + 2 = x (c+1)\)

You can plug in number of candies in the equation all of them will give us an integer for x (the number of children) EXCEPT FOR answer choice D the logic here is that we cannot have half a child? :grin:

if we substitute the number candies to be 1, we get number of children 64/2 = 32

if we substitute the number candies to be 3, we get the number of children 64/4 = 16

If we substitute the number candies to be 7, we get the number of children 64/8 = 8

If we substitute the number of candies to be 9, we get the number of children 64/10 = 6.40 not possible!

if we substitute the number of candies to be 15, we get the number of children to be 64/16 = 4

Answer choice D

I hope this helps!
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Re: 62 candies were equally distributed to a group of children. If the num  [#permalink]

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New post 16 Oct 2018, 14:46
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Top Contributor
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15


There's a nice rule that says, If N divided by D equals Q with remainder R, then N = DQ + R
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

Let n = number of children
Let c = number of candies that each child received.

So, we can rephrase the given information as: When 62 candies are divided among n children, each child receives c candies and there are (n-2) candies remaining.
In other words, 62 divided by n equals c with remainder (n-2)

Applying the above rule, we get: 62 = cn + n-2
Add 2 to both sides to get: 64 = cn + n
Factor out the n to get: 64 = n(c + 1)
Divide both sides by n to get: 64/n = c + 1
Subtract 1 from both sides to get: c = 64/n - 1

This tells us that c is 1 less than some divisor of 64 (64/n must be a divisor of 64)

Now check the answer choices.

a. 1. Since 2 is a divisor of 64, we can see that 1 is 1 less than some divisor of 64. ELIMINATE A
b. 3. Since 4 is a divisor of 64, we can see that 3 is 1 less than some divisor of 64. ELIMINATE B
c. 7. Since 8 is a divisor of 64, we can see that 7 is 1 less than some divisor of 64. ELIMINATE C
d. 9. 9 is NOT is 1 less than some divisor of 64. KEEP D
e. 15. Since 16 is a divisor of 64, we can see that 15 is 1 less than some divisor of 64. ELIMINATE E

Answer: D

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Re: 62 candies were equally distributed to a group of children. If the num  [#permalink]

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New post 16 Oct 2018, 15:27
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15

\(\left. \matrix{
{\rm{\# children}}\,\,\,{\rm{:}}\,\,\,n \ge 2\,\,{\mathop{\rm int}} \hfill \cr
{\rm{\# candies/child}}\,\,\,{\rm{:}}\,\,\,c \ge 1\,\,{\mathop{\rm int}} \,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,nc\,\,\,{\rm{candies}}\,\,{\rm{distributed}}\)


\(?\,\,\,:\,\,\,c\,\,\,\underline {{\text{CANNOT}}}\)

\(62 - nc = n - 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,n\left( {c + 1} \right) = 64\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{FOCUS}}\,!} \,\,\,\,c + 1\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{positive}}\,\,{\text{divisor}}\,\,{\text{of}}\,\,64\,\,\,\,\left( * \right)\)

\(?\,\,\mathop = \limits^{\left( * \right)} \,\,\left( D \right)\,\,\,\,\,\,\,\left[ {\,10\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{a}}\,\,{\text{positive}}\,\,{\text{divisor}}\,\,{\text{of}}\,\,64\,} \right]\,\,\,\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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62 candies were equally distributed to a group of children. If the num  [#permalink]

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New post 16 Oct 2018, 18:49
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15


let c=number of children
x=number of candies received by each child
x=[62-(c-2)]/c→
x=(64-c)/c
plugging in all answer choices for x,
only 9 makes c a non-integer
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Re: 62 candies were equally distributed to a group of children. If the num  [#permalink]

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New post 16 Oct 2018, 20:00
1
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15



Hi..

A good method would be to make use of choices here
First let us make the equation..
Let x be the number of candies given to each of n child..
So left is n-2..
Therefore n*x+n-2=62......n (x+1)=64
Now the choices..
a.1......n(1+1)=64....n=32
b.3.....n(3+1)=64...n=16
c.7.....n(7+1)=64...n=8
d.9.....n(9+1)=64..10n=64...n=6.4 Not possible as number of children cannot be a fraction
e.15....n(15+1)=64...n=4

D
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Re: 62 candies were equally distributed to a group of children. If the num   [#permalink] 16 Oct 2018, 20:00
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