How many apples did he start with?

Question

A man had to pack a sack of apples into packets but as each
packet had to have exactly the same number of apples he was
having difficulty.
If he packed 10 apples per packet, one packet only had 9.
If he packed 9 apples per packet, one packet only had 8.
If he packed 8 apples per packet, one packet only had 7.
If he packed 7 apples per packet, one packet only had 6.
If he packed 6 apples per packet, one packet only had 5.
If he packed 5 apples per packet, one packet only had 4.
If he packed 4 apples per packet, one packet only had 3.
If he packed 3 apples per packet, one packet only had 2.
If he packed 2 apples per packet, one packet only had 1.

How many apples did he start with?

Please explain how you calculated your answer.

Answer 1

Pretty straightforward.

Answer is one less than the smallest number that contains all the factors.

So start with the factors
10 = 2 X 5
9 = 3 X 3
8 = 2 X 2 X 2
7 = 7
6 = 2 X 3
5 = 5 (subset of 10)
4 = 2 X 2 (is a subset of 8)
3 = 3 (contained in 9 and 6)
2 = 2 (contained in 4, 6, 8, and 10)

So let's pull em out. There are three 2's, two 3's, one 5, and one 7.

So we get 2 X 2 X 2 X 3 X 3 X 5 X 7 = 2520

Subtract 1 from that (since we always are one short of a full packet) and we get 2,519!

Tada!

Hope that helps!

Answer 2

If the man had just one more apple, no matter how many apples he had packed per packet, it would come out evenly (with 2-10 per packet). Hence, the number of apples he started with is 1 less then a number that's evenly divisible by 2, 3, 4, 5, 6, 7, 8, 9, and 10. There's an infinite number of answers to this question, with the simplest one to solve would be all the digits 2 though 10 multiplied by each other minus 1, or 10!-1, or 3,628,799.

However, you're probably looking for the smallest number of apples which would work. If a number is evenly divisible by 8, then it's automatically divisible by 2 and 4. Similarly, a number evenly divisible by 9 it's already divisible by 3 to begin with. And if a number's divisible by both 8 and 9, it'll be divisible by 6. So what you're looking for is 5*7*8*9 - 1 = 5*8*7*9 - 1 = 40*63-1 = 2520-1 = 2519.

Answer 3

Give him one more apple. Then, no problem, the total number of apples will be divisible by 2, 3, . . .9. The least of the numbers divisible by 2, 3, 4, . . . 9 is:

2^3 * 3^2 * 5 * 7 = 2520

So the man has 2520*k - 1 apples where k an integer >= 1.
For k = 1 he has 2519 apples
For k = 2 he has 5039 apples
and so on . .

Answer 4

1

Answer 5

Since each packing reveals that the number of apples modulo the packing number was -1, then the easiest way for this to happen is to have 10!-1 apples to begin with.
10!-1 = 3628799 and
3628799/10 = 362879 remainder 9
3628799/9 = 403199 remainder 8

etc.

(it's too early for me to be figuring out the prime factors of everything from 2 thru 10 ☺)


Doug

Answer 6

they are happening the rest room. i in my opinion am NO fan of apple. they don't seem to be very sturdy on the 2nd. the reason they are nevertheless doing decently is with the help of there committed iSheep. besides the fact that, they have an exceedingly gloomy destiny in the event that they do no longer shape up and grant better products at better costs.

Answer 7

What number is one less than the one which is divisible by 9,8,7 and 5? It would be 2520-1 = 2519.

Answer 8

I am really good at math. I think he started with 2519 apples. Sorry if it's wrong.

Answer 9

90

Answer 10

To hell with the apples. Your good-looking.