Question

Tracy had a bag of candies, and none of the candies could be broken into pieces. She ate $\frac{1}{3}$ of them and then gave $\frac{1}{4}$ of what remained to her friend Rachel. Tracy and her mom then each ate 15 candies from what Tracy had left. Finally, Tracy's brother took somewhere from one to five candies, leaving Tracy with three candies. How many candies did Tracy have at the start?

Answer 1

72

Explanation:

Let x be Tracy's starting number of candies. After eating
`$(1)/(3)$` of them, she had
`$(2)/(3)x$` left. Since
`$(2)/(3)x$` is an integer, x is divisible by 3. After giving
`$(1)/(4)$` of this to Rachel, she had
`$(3)/(4)$` of
`$(2)/(3)x$` left, for a total of
`$(3)/(4) \cdot (2)/(3)x = (1)/(2)x$`. Since
`$(1)/(2)x$` is an integer, x is divisible by 2. Since x is divisible by both 2 and 3, it is divisible by 6.

After Tracy and her mom each ate 15 candies (they ate a total of 30), Tracy had
`$(1)/(2)x - 30$` candies left. After her brother took between 1 and 5 candies, Tracy was left with 3. This means Tracy had between 4 and 8 candies before her brother took some candies. Hence,

`4 \le (1)/(2)x - 30 \le 8\qquad \Rightarrow \qquad 34 \le (1)/(2)x \le 38\qquad \Rightarrow \qquad 68 \le x \le 76.`

Since x is divisible by 6, and the only multiple of 6 in the above range is 72, we have x =
`\boxed{72}`.

Answer 2

72

Explanation:

Let the number of candies = n

Tracy ate = 1/3 of n = n/3 remaining 2n / 3

she gave 1/4 of the remainder to her friend = 1/4 of ( n - n/3) = 2n / 12

the new remainder = (2n / 3) - (2n / 12) = n /2

she and her mom then ate altogether = 30

and the brother took from 1 to five, the number he took = x

and she has 3 left

( n/2) - 30 - x = 3

n = 2 ( 33 + x)

now we know that the candies could not be broken and x is between 1 to 5

n = 2 (33 + 1), 2 (33+2), 2(33 +3), 2 (33+ 4) and 2 (33 + 5) this are the possible values of n, ( 68, 70, 72, 74, 76) and the multiple of 3 is 72

n therefore = 72