Question

Clara has some sweets. If she packs them in groups of 5, she will have 3 extra sweets. If she packs them in groups of 8, she will have 7 extra sweets. What is the minimum number of sweets Clara has?

Answer 1

we'll first lighten the load of integer substitution with modulus arithmetic.
\(a \equiv 3\) (mod 5)
so, \(a+2 \equiv 5\) (mod 5)=> \(a+2 \equiv 0\) (mod 5)
similarly for \(a \equiv 7\) (mod 8), \(a+1 \equiv 0\) (mod 8)
so, \(\frac{a+2}{5}=x\), and \(\frac{a+1}{8}=y\) where x,y are integers.
so, the integer number a, must be a odd number(because a+1 must be even), and should have 3 at it's unit because only then the a+2 might be a multiple of 5.
and following that, 3+1 will cause a 4 as digit in the\(\frac{a+1}{8}=y\) and the multiples of 8 that have 4 are 24, 64, 104,...
we'll try the first, for \(a+1=24\), then a=23.

subbing that into\(\frac{a+2}{5}=x\) gives an integer answer for x.
Thus a=23.
do you need further help?

Answer 2

Thank you for responding. although this was for my 10 year old son and I was wondering if there is an easier (non-algebraic) approach such that it would be easier for him to understand. I appreciate the help.

Answer 3

well, in that case, he'll have to substitute the numbers.
First case: the number taken out 3 is divisable by 5.
Second case: the number subtracted by 7 is divisible by 8.

thus we first list the multiples of 5:
5,10,15,20,25,30,35...

we try to add 3 from each multiple(reversing the thought. As the number taken out 3 is a multiple, the multiple added 3 is the number)

8,13,18,23,28,33,38.....
now these may be candidates for the number. We then subtract 7 from each candidate to see if they can follow the second case.

1,6,11,16,21,26......

it is obvious that 16 is divisible by 8. so, 23 is the number.
you can also start from listing the multiples of 8.

Answer 4

Thanks so much! we also finally figured out what to do, added 3 and 7 to the multiples of 5 and 8 and saw that 23 ended up as the common number between the multiples. But thanks really, we appreciate the help!

Answer 5

you're welcome :)