Puzzle:

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. And so on down to 2 apples. How many apples did he start with?

Solution:

2519 apples.

The solution for the answer is the LCM (Lowest Common Multiple) of 10,9,8,7,6,5,4,3,2,1 minus 1. LCM would give the least number which is divisible by all of these number and subtracting one would give us the number of apples which were initially there.

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Solution:

2519 apples.

The solution for the answer is the LCM (Lowest Common Multiple) of 10,9,8,7,6,5,4,3,2,1 minus 1. LCM would give the least number which is divisible by all of these number and subtracting one would give us the number of apples which were initially there.

## 22 comments:

2519

Well, Chavali, your answer is correct...

It is 2519...

But I dont know how you did it...It'd be nice if people could also give a brief sketch as to how they went about it...

More so because these are supposed to be solved like puzzles, and not by brute force (using the computer)...

So here's how I went about solving it...I used the method of construction...Basically, start from N=2 and go on till N=10...

N = no. of fruits in each case

P(n) = total no. of fruits for N = n

1. N = 2... every odd number case,

so lets take P(2) = 3

2. N = 3... Now we need to add a few more fruits, so that P%3 = 2, but we must also add properly, so that the previous cases are still valid (here only N = 2 case)...

so we have,

P(3) = P(2) + X...

Now check what remainder remains when we divide P(2) with 3...in this case, its 0, so we need that

X%3 == 2 and also X%2 == 0...

So best is, choose X = LCM(all previous cases) * Y, such that X%3 = 2...

So, we have X = 2*1 = 2

therefore, P(3) = 5...

iii. N = 4

similarly,

P(4) = P(3) + X

= 5 + LCM(2,3)*Y

and 5%4 = 1, therefore we just need X%4 = 4- (P(3)%4)

so, Y = 1, ie, X = 6

therefore P(4) = 5 + 6 = 11...

following on, we will get

iv. N = 5, P(5) = 11+48 = 59

v. N = 6, P(6) = 59+60 = 119

vi. N = 7, P(7) = 119+300 = 419

vii. N = 8, P(8) = 419+420 = 839

viii. N = 9, lets do this now incase you didnt get it...

P(9) = P(8) + X

X = LCM(2,3,4,5,6,7,8)*Y

such that X%9 == (8-P(8)%9)

i.e. X = 1680

therefore, P(9) = 2519

which is what we need...because it satisfies for N = 10 as well :)...

answer is 2519

simple, lcm of 2,3,4,5,6,7,8,9,10 is 2520 and

so 2520 -1 will give remainders 1,2,3,4,5,6,7,8,9 when divided by 2,3,4,5,6,7,8,9,10

Yes..that's the procedure

Damn... Y didnt I think of that :)...

Thanks PG, owe you big time :)...

THERE CAN BE MANY ANSWERS...2519 ARE THE LEAST NUMBER OF APPLES..KEEP ON ADDING 2520 TO 2519...N U'LL BE GETTING MORE N MORE ANSWERS..

I figured out by taking all numbers from 1 to 10, and leaving out the combinations which are already taken care by the number considered.

So the reqd no will then be 5*7*8*9-1=2519. All other nos can be accounted for within this prod itself! (For eg, 10=2*5, 2 is accounted for in 8).

ummmh u guys are making it too difficult on yourselves... simple answer is 89... you guys dont know your LCM

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